The author of these lessons is Dr. Thomas G. Cleaver, Professor Emeritus of Electrical Engineering, The University of Louisville. The author wishes to acknowledge the contributions of Dr. Joseph D. Cole and Dr. William H. Pierce.

Below is a summary of the Code of Professional Practice and Conduct for professional engineers and land surveyors in Kentucky [Kentucky Engineer, Vol. 37, No. 8, Sept. 2001].

The engineer ... shall conduct his practice in order to protect the public health, safety, and welfare.

A licensee shall issue all professional communications and work products in an objective and truthful manner. A licensee shall be objective and truthful in all professional reports, statements or testimony and shall include all material facts.

A licensee shall avoid conflicts of interest.

A licensee shall not knowingly associate with any person engaging in fraudulent, illegal or dishonest activities. A licensee shall not aid or abet the illegal practice of engineering.

A licensee shall perform his services only in the area of his competence.

The professional engineer ... shall avoid conduct likely to discredit or reflect unfavorably upon the dignity or honor of his profession.

Professional ethics and personal ethics are really no different. When you
graduate as engineers, you will be expected to practice your profession with
honor and integrity. This means that you must tell the truth, never take
credit for the work of another, treat others with respect, and give your
employer good value for your salary. As a student, your role is very similar.
You tell the truth; you do not copy homework or cheat on exams; you treat your
classmates and your instructors with respect; you work hard at your classes.
For your instructor, there is **nothing** more important than your growth as
people of strong character.

You are expected to know the International System (SI) of units (the metric system), because units in electricity are based on this system. For some practice with SI units, click on this game.

Note that current is defined as the flow of **positive** charges.

v=Ri is Ohm's Law. It's an important equation; commit it to memory.

Mega (10^{6}), kilo (10^{3}), milli (10^{-3}), micro
(10^{-6}), nano(10^{-9}), and pico(10^{-12}) are the
most frequently used prefixes. Memorize them.

The basic power equation is p = iv. Lower case is used to show that this
equation is true for time-varying voltages and currents. If you know the
current through a resistor, use p = i^{2}R. If you know the voltage
across a resistor, use p = v^{2}/R.

Two voltage sources are shown below.

The one on the left is the standard representation for a DC voltage source. The
one on the right is, technically, the symbol for a battery. For our purposes,
we can consider both symbols to represent general, ideal, DC voltage
sources.

Be sure you understand the definitions of branch, node, and loop.

Note the subscripting convention. v_{ab} is positive if point a is
more positive than b.

Lower case letters (v, i) indicate time-dependent quantities.

Upper case letters (V, I) indicate time-independent quantities, such as DC values.

A voltage source has the property that its voltage is independent of the current through it. Thus, a "6 Volt battery" will have 6 volts across its terminals whether it is connected to a 1 Ω resistor, or to a 1 million Ω resistor. It should be understood that what we are talking about here is an "ideal" voltage source. The voltage of a real voltage source will depend somewhat on the current through it. Unless otherwise specified, we will always be using ideal elements. Also note that current can flow either way through a battery. It supplies energy when current flows out of the positive end; it absorbs energy ("charging" the battery) when current flows into the positive end.

An ideal current source has the property that its current is independent of the voltage across it. A "5 amp source" will supply 5 amps to a short circuit with zero voltage or 5 amps to a 1 million Ω resistor causing a 5 million volt drop. The voltage across a current source (both magnitude and direction) is always unknown until the rest of the circuit is defined.

Although Kirchhoff's Laws will work for any direction chosen, for consistency, we will always go around loops in the clockwise direction, and we will count voltage drops as positive.

Click here to see a demonstration of how voltage, resistance, and current are related.

Watch your signs when doing Kirchhoff's Laws problems. This is the principal source of student errors.

Nodes that are connected by a wire (a short circuit) can (and should) be treated as a single node.

Before going on, you should complete Tutorial 1A on **the
physics of electricity**.

*This section was written by William H. Pierce.*

A voltage generator imposes its stated voltage as the **difference** in
voltage between its two terminals. In many cases, that does not mean that the
voltage to ground is the stated value of the generator. In the circuit below,
for example, V_{a} is not 2 V, nor is V_{b} 2 V or -2 V. What
is true is that

V_{a} - V_{b} = 2 V.

We can solve that equation to get

V_{a} = V_{b} + 2 V.

Of course, if the minus terminal were at the top, it would be

V_{a} = V_{b} - 2 V.

In Ohm's Law, I = V/R, the voltage V must be the **difference** in
voltage of the two terminals of the resistor. The current arrow points into the
positive end of a resistor, and comes out the negative end. This also
determines which terms in the Ohm's Law equation we take as positive and
negative. In this example,

I = (V_{a} - V_{b})/R.

Notice that V_{a}, the first term, is associated with where the
current comes into the resistor, and V_{b}, the second term, is
associated with where the current comes out.

If V_{a} = 500 V, V_{b} = 600 V, and R = 10 Ω, then I
will be -10 A. Yes, that's **minus** 10.

Remember that the laws for a single component only apply to the difference of the voltages on its terminals.

A current is defined by its location in the circuit, and an arrow indicating the direction in which it is defined. A current arrow hitting a resistor creates a voltage difference that is IR, with the positive sign of this difference defined on the side the arrow first hits. It will be helpful to actually write a + sign on the side the arrow first hits, in order to avoid mistakes. This arrow and + sign technique should also be used for inductors and capacitors.

In finding the voltage of a point when only differences are known, begin at
ground (0 volts) and take a sum of the signed numbers using the sign on the
**far** side of the difference. (The far side is the one closer to the point
and farther from ground). Consider the circuit below.

To find V_{a}, begin at ground, add (+3 V) for the battery on the
left, add (+2 × 4 V) for the contribution from the 4 Ω resistor, add (-4
× 5 V) for the contribution from the 5 Ω resistor, and add (-7 V) for the
contribution from the battery on the right. This gives

V_{a} = 3 + 8 - 20 - 7 = -16 V.

Before going on to the homework, you should complete Tutorial 1 on **basic
circuit laws**.

**Important:** Sketch each circuit. Label all circuits with appropriate
currents and voltages.

For the first three problems, there are no "right" or "wrong" answers. You will be given credit as long as your answers are thoughtful and sincere.

- You meet with three of your classmates at the library to work together on
the calculus homework. Do you think this helps you learn the material
faster or better? How strong is the temptation to copy a solution, rather
than to learn how to solve the problem? Where does one draw the line
between copying and getting help?

- During an exam, you see a student pass a note to another student. What
do you do?

- Someone a year ahead of you in engineering offers to give you all his old
homeworks and labs. You know your teacher frequently uses problems and
labs from previous terms. What do you do?

- How much energy does it take to move an electron from the negative
terminal of a 1.5 V battery to the positive terminal? Be careful of your signs.

- E = 36 V, R1 = 8 Ω, R2 = 4 Ω. Find the current through R1
and the power consumed by R2. Do not solve by combining resistors in
series. Use Kirchhoff's Laws. The answers are integers. Hint: Define
voltages across the two resistors. Write one KVL equation and two Ohm's Law
equations.

- E = 500 V. All the resistors shown are identical. Each one consumes 400
W. Give the value of any one resistor. Do not solve by combining resistors
in series. Use Kirchhoff's Laws. The answer is an integer. Hint: Define
voltages across the resistors. Write a KVL equation.

- E = 72 V, R1 = 60 Ω, R2 = 40 Ω. Find I. Do not solve by
combining resistors in parallel. Use Kirchhoff's Laws. The answer is an
integer. Hint: Define currents in the two resistors. Write two KVL
equations, one KCL equation, and two Ohm's Law equations.

- E = 200 V, R1 = 8 Ω, R2 = 30 Ω, R3 = 20 Ω. Do not
combine resistors in series and parallel. Using Kirchhoff's Laws, find the
current in each resistor. The answers are integers. Hint: Define voltages
and currents. Write one KCL equation, two KVL equations, and three Ohm's
Law equations.

- E1 = 12 V, E2 = 20 V, R1 = 8 Ω, R2 = 3 Ω, R3 = 15 Ω.
Find the voltage at point A with respect to ground. The answer is an
integer.

**Bonus ** (no partial credit). It is wise to become adept at
solving simultaneous equations using modern tools. Advanced calculators have
this capability, as do computer algebra systems, such as Maple. Use either a
calculator with equation-solving capability or a computer algebra program to
find V_{1}, V_{2}, V_{3}, and V_{4} in the
following equations:

6V_{1} - 12 V_{2} + 8V_{3} = 11

4V_{1} - 9V_{2} + 15V_{3} - 7V_{4} = -19

2V_{2} - 3V_{3} + V_{4} = 5

-10V_{1} - 13V_{2} + 14V_{3} + 6V_{4} = 0

Describe the tools you use and how to use them to solve this problem.

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Note that current is defined as the flow of **positive** charges.

v = Ri is Ohm's Law. It's an important equation; commit it to memory.

Click here (Florida State University) to find out how to use the resistor color code.

Mega (10^{6}), kilo (10^{3}), milli (10^{-3}), micro
(10^{-6}), nano(10^{-9}), and pico(10^{-12}) are the
most commonly used multipliers. Commit them to memory.

Two elements are in series if they are connected together at one end with no other connection at that end. Use this definition, rather than your intuition, to determine if elements are in series. The following elements are in series:

These elements are not in series:

For resistors in series, the net resistance is just the sum of the individual resistances.

R_{EQ }= R_{1} + R_{2} + ..... R_{n}

Two elements are in parallel if both ends of each element are connected together. These elements are in parallel:

These elements are not in parallel:

A parallel combination of resistors is found by the equation

1/R_{EQ} = 1/R_{1} + 1/R_{2} + 1/R_{3} +
..... 1/R_{n}

A useful special case is for exactly two resistors in parallel:

R_{EQ} = R_{1}R_{2}/(R_{1} + R_{2})

The equations for elements in series and parallel are easily derivable from Kirchhoff's Laws.

Before going on, you should complete Tutorial 2 **on resistors
in series and parallel**.

Some circuits, such as the one shown below, cannot be simplified by combining elements in series and parallel. When this happens, you just have to grit your teeth and apply Kirchhoff's Laws, or use the delta-wye transformation (discussed in a later section).

An open circuit is a place in a circuit where nodes are not connected, or open. Zero amps flows between nodes that are not connected, meaning zero amps flows in an open circuit. The resistance across an open circuit is equal to infinity. Open circuits are represented as a broken wire. For calculating an equivalent resistance, a resistor connected to the circuit at only one node is open. An open resistor (1) makes zero Ohms of contribution to the equivalent resistance and (2) can be removed from the circuit when calculating the equivalent resistance.

Open circuit = 0A of current

Open circuit = ∞Ω of resistance

An element (e.g., resistor, voltage source, etc.) is shorted if both of its ends are connected to the same one node. Short circuits are represented as a wire. A wire is considered to have a negligible amount of voltage, or zero volts, meaning the voltage is zero for a short circuit. The resistance of a wire in electrical circuits is considered to be negligible, or 0Ω. Therefore, the resistance across a short circuit is negligible, and considered equal to zero. For calculating an equivalent resistance, a shorted resistor is one whose both ends are connected to the same one node. A shorted resistor (1) makes zero Ohms of contribution to the equivalent resistance and (2) can be removed from the circuit when calculating the equivalent resistance.

Short circuit = 0V of voltage

Short circuit = 0Ω of resistance

The voltage divider equation will be very useful to you. Consider the figure below.

It is easily derivable from Kirchhoff's Laws that

V_{2} = V_{S}R_{2}/(R_{1} + R_{2})

The current divider equation may be occasionally useful. Consider the figure below.

Similar to the voltage divider equation, Kirchhoff's Laws can be used to find that

I_{2} = I_{S}R_{1}/(R_{1} + R_{2})

Notice that the numerator term uses the resistor that the current
**doesn't** go through.

Before going on to the homework, you should complete Tutorial 2A on **voltage
and current dividers**.

Please note: It is not necessary to use delta-wye transformation in any of these problems.

- R1 = 11 Ω, R2 = 15 Ω, R3 = 30 Ω, R4 = 2 Ω. Find
the resistance between X and Y. The answer is an integer.

- R1 = 42 Ω, R2 = 80 Ω, R3 = 120 Ω, R4 = 45 Ω.
Find the resistance between E and F. The answer is an integer.

- R1 = 6 Ω, R2 = 9 Ω, R3 = 15 Ω, R4 = 14 Ω, R5 = 10
Ω, R6 = 30 Ω, R7 = 2 Ω. Find the resistance between A
and B. The answer is an integer.

- Using exactly six 10 Ω resistors and no other resistors, design and sketch a circuit with a resistance of exactly 22 Ω. All six resistors must be significant parts of the circuit.
- a) Design and sketch a circuit with a resistance of exactly 1.4 MΩ
between nodes A and B. Use only the following list of resistor values in
your design: 100 kΩ, 620 kΩ, 2.4 MΩ, and 3.3 MΩ .
Use as many of the listed resistor values as you choose for your design.
Use no other resistor values. All resistors used must be a significant part
of the circuit.

b) Go to digi-key.com. Search for resistors to build your design from part a). Use search filters for resistors with mounting type “Through Hole,” a power rating of “1/4W (.25W),” a material composition of “Carbon Film,” a resistance value tolerance of “±5% (directly above ‘Jumper’ listing in Tolerance),” from the manufacturer “Yageo,” and in packaging of type “Bulk.”

c) What would it cost (Subtotal) to build 10 versions of your above design? (Click on each ‘Digi-Key Part Number’ and add specific Quantities to your order.)

d) What would it cost (Subtotal) to build 10,000 versions of your above design?

e) Consider the point of view of a resistor manufacturer. Why would a manufacturer not stock a 1.4 MΩ resistor but would stock 1.3MO and 100 kΩ resistors? - Sketch a voltage divider circuit that uses a battery of your choice, a 47 kΩ resistor, and a 22 kΩ resistor to produce an output voltage of approximately 6.13 V.
- The 6.8 kΩ resistor has a tolerance of 10%. The 3.3 kΩ
resistor has a tolerance of 20%. What is the maximum possible voltage for
Vout? What is the minimum possible voltage for Vout?

- A student uses a voltage divider in hopes of converting 9 V to 1 V, using
the circuit shown. When the student connects a cheap volt-ohm meter (VOM)
to Vout, she gets just 0.67 V. Explain why the voltage is lower than
expected. Note: This problem has nothing to do with tolerance; the
resistor values are accurate.

**Bonus **(no partial credit). All resistors are 1 Ω. Find an
expression for R_{GH} that can be expanded to as many decimal places as
desired, *e.g.*

R_{GH} = (π - 3)/7 (Not the right answer).

Before going on to the node-voltage and mesh-current methods of solving circuits, you must thoroughly understand how to solve circuits by applying Kirchhoff's Laws.

To completely "solve" a circuit, we must know the voltage across and the current through each element. The technique is as follows:

- Label the circuit:
- Assign a current (showing its direction) in every element. Elements in the same branch should be assigned the same current. Don't forget to assign currents to voltage sources.
- Assign a voltage across every element. If it's a resistor, inductor, or capacitor the + sign should be placed where the current enters the element. Don't forget to assign voltages to current source.
- Identify each essential node (where 3 or more wires join). The node need not be a single point; it may stretch across a circuit and go around corners. Label each node with a letter of the alphabet.
- Identify the simple loops (meshes). These are the "window panes." They are loops that don't contain other loops. Label each one with a number.

- Write an Ohm's Law equation for each resistor. V = IR.
- Count your nodes. Decide which node to ignore (usually the most complex one). Write a Kirchhoff's Current Law equation for each of the remaining nodes. Take the currents leaving the node as positive.
- Write a Kirchhoff's Voltage Law equation for each mesh. Go clockwise around each mesh and take voltage drops as positive. Don't forget voltages across current sources.
- Count your equations and unknowns. They should be equal.
- Solve the set of simultaneous equations.

We will employ the process above to completely solve the circuit shown in Figure 1 below.

Figure 1. Circuit to be solved.

Step 1a is to assign currents to every element. This is shown in Figure 2, below.

Figure 2. Circuit with currents defined.

The current in the 6 V battery and the 9 Ω resistor is already known
to be 3 A, so it is not necessary to define a current in that branch. Note the
source in the upper branch. This is a *dependent voltage source*. Its
voltage is dependent on the current in the center branch. Although it is
dependent on a current, it is still a voltage source, not a current source. It
is therefore necessary to define a current in the branch with the dependent
source. The direction of current I_{8} is *arbitrary*. The
current may be defined in either direction.

Step 1b is to assign voltages, as shown below in Figure 3.

Figure 3. Circuit with currents and voltages defined.

For each of the resistors, a voltage is assigned. The polarity of each voltage is chosen with the + sign where the current comes in. A voltage must also be assigned to the 3 A current source. The polarity of this voltage is arbitrary.

Step1c requires that essential nodes be identified circled and labled. This is shown below in Figure 4. Step 1d is to identify and label meshes. This is also done in Figure 4.

Figure 4. Circuit with all necessary labels.

Now that the circuit is completely labeled, it's time to write the equations. Step 2 calls for writing the Ohm's Law equations. We write one for each resistor. These are shown below.

V_{4} = 4I_{4}

V_{8} = 8I_{8}

V_{9} = 3×9

In step 3, we write the Kirchhoff's Current Law Equations. There are two essential nodes. We write one fewer equation than there are nodes, so we need only one equation. We can write it at either node A or node B. Let's use node B:

3 + I_{4} - I_{8} = 0

In step 4, we write Kirchhoff's Voltage Law Equations. There are two meshes. We write one equation for each mesh:

Mesh 1: V_{8} - 5I_{4} + V_{4} = 0

Mesh 2: V_{9} - V_{4} + 6 - V_{3} = 0

In step 5, we count the equations and unknowns. There are 6 equations and 6
unknowns (V_{3}, V_{4}, V_{8 } , V_{9},
I_{4}, I_{8}), therefore we can be confident that we can solve
the circuit.

Solving the circuit (step 6) yields the following results:

V_{3} = 46.714 V

V_{4} = -13.714 V

V_{8} = -3.429 V

V_{9} = 27 V

I_{4} = -3.429 A

I_{8} = -0.429 V

Most of the results were negative. Choosing the other direction for
I_{8} would have resulted in additional positive values, but that
doesn't matter.

Step 6, above, "Solve the set of simultaneous equations," can be greatly simplified by use of modern tools. High-end calculators can solve sets of simultaneous equations. If you have such a calculator, you should learn to use this feature. Computer algebra systems such as Maple can also be used to solve simultaneous equations.

Below are five links to simulations of Kirchhoff's Laws. They were created
by Sergey Kiselev and Tanya Yanovsky-Kiselev.

Single
loop

Double
loop with three sources

Double
loop with three resistors

Double
loop with three resistors and two sources

Double
loop with three resistors and three sources

Here's another link to a site that allows you to design and build a circuit with resistors, light bulbs, ammeters, voltmeters, etc. It was created by the Article 19 Group. It requires the Shockwave plugin.

Before going on to the homework, you should complete Tutorial 3 on **circuit
solving with Kirchhoff's Laws**.

**Note: **If you use a simultaneous-equation-solving
calculator or a computer algebra system to solve a homework problem, note this
fact in your solution. This applies for all homework problems, not just the
ones in this lesson.

- Completely label the diagram. Use Ohm's Law and Kirchhoff's Laws to write
enough equations to solve for the current through and the voltage across
each element. Do not solve. Note: The purpose of
this exercise is to give you practice in carefully labeling circuits and
writing the circuit equations. Caution: There is a dependent source in
this problem; is it a voltage source or a current source? Another
caution: Make sure your current directions match your voltage polarities
for resistors.)

- Label the diagram as appropriate. Write enough equations to solve for the
unknown V
_{4}. Do not solve. Do not include more equations than you need. The purpose of this exercise is to help you recognize the features of a circuit that are not necessary for a solution. Hint: Write the equations to completely solve the circuit, then cross out the ones that aren't needed. Further hint: If an unknown appears in but one equation, and that unknown is not needed, the equation (and unknown) can be eliminated. (Caution: There is a current source in this problem. Does a current source have a voltage across it?)

- E = 3I, R1 = 8 Ω, R2 = 7 Ω, Is = 6 A. Find I. The answer is
an integer. (Caution: There is a dependent source in
this problem. Is it a voltage source or a current source?)

**Bonus **(no partial credit).** **Find the current I in the 10
Ω resistor. The answer is not an integer.

The node voltage method simplifies the writing of equations for a circuit because you need only write Kirchhoff's Current Law. Kirchhoff's Voltage Law is implicit in the writing of the other equations. Consider the resistor below.

The current is given by i = v/R by Ohm's Law. But v is v_{ab} which
is v_{a} - v_{b}, so

i = (v_{a} - v_{b})/R.

Stated in words, this is: "The current flowing from *a *to *b* is
the voltage at *a *minus the voltage at*b* divided by the
resistance". Similarly, we can say the current that flows from *b* to
*a* is

(v_{b} - v_{a})/R.

We don't have to know in advance which voltage has a bigger magnitude in order
to choose which current we want (*a* to *b* or *b *to
*a*) because the sign of the answer will take care of that, and

i_{ab} = -i_{ba}, just as

v_{ab} = -v_{ba}.

Using Ohm's Law we can write Kirchhoff's Current Law at any node in terms of the voltage at the nodes. Note that it is not necessary to write an equation at every node. You never write an equation at the reference node (this is, by definition, zero volts), and it is not necessary to write an equation at a node where the voltage is known.

An example is given below.

First, we must identify the nodes. Note that the line between the 7 Ω resistor and the 3 Ω resistor is one node. There are 4 nodes. One of those must be chosen as the reference node. A good choice is the one at the bottom. This is convenient because it gives us an immediate answer for the node on the right (5 V). Now we can label all the nodes:

We can immediately assign 5 V to the node on the right, because we observe that it is 5 V higher than the reference node.

The "node" at the upper right labeled "V_{B} + 9" is a non-essential
node. It is labeled for convenience. We see that the voltage at this point is 9
V higher than V_{B}.

We now write KCL at the remaining two nodes:

-2 + (V_{A} - V_{B})/3 + (V_{A} - V_{B})/6 +
(V_{A} - 5)/7 + V_{A}/4 = 0

(V_{B} - V_{A})/3 + (V_{B} - V_{A})/6 +
(V_{B} + 9 - 5)/8 = 0

When writing these equations, we assume that the current out of each node is
positive. In the first equation, this gives us terms like:

(V_{A} - V_{B})/3,

and in the second, we have:

(V_{B} - V_{A})/3.

This is not inconsistent, because

(V_{A} - V_{B})/3 = -(V_{B} - V_{A})/3.

The term

(V_{B} + 9 - 5)/8

will take a little explanation. We want a term for the current between the
nodes labeled V_{B} and 5 V. The same current flows through the 9 V
battery as flows through the 8 Ω resistor. Therefore, if we can determine
the current through the 8 Ω resistor, we will know the current through
the 9 V battery. Now it should be clear why it was convenient to label the
"node" at the upper right as V_{B} + 9. Using i = v/R, the current
through 8 Ω resistor is

[(V_{B} + 9) - 5]/8.

We have two equations and two unknowns. Solve by any method you wish. Once way is shown below.

2 = (V_{A} - V_{B})/3 + (V_{A} - V_{B})/6 +
(V_{A} - 5)/7 + V_{A}/4

2 + 5/7 = V_{A}(1/3 + 1/6 + 1/7 + 1/4) - V_{B}(1/3 + 1/6)

2.714 = .893V_{A} - .5V_{B}

0 = (V_{B} - V_{A})/3 + (V_{B} - V_{A})/6 +
(V_{B} + 9 - 5)/8

-4/8 = -V_{A}(1/3 + 1/6) + V_{B}(1/3 + 1/6 + 1/8)

-.5 = -.5V_{A} + .625V_{B}

The result is

V_{A} = -1.79 V

V_{B} = -2.23 V

Take care to avoid round-off error when working with decimals. Some people prefer to work with fractions to avoid this problem.

- Identify (mark) the essential nodes.
- Choose a reference node (usually the node with the most branches connected to it).
- For each voltage source, write an appropriate equation, or define the voltage at a point on the circuit. It may be necessary to define a current in some voltage sources.
- Assign a variable name to each other essential node.
- Write Kirchhoff's Current Law for every node (except the reference node and nodes where the voltage is known) using Ohm's Law.
- Count your equations and unknowns (they should be equal).
- Solve.
- Is your answer reasonable in terms of magnitude, sign, and units?

Before going on to the homework, you should complete Tutorial 4 on **the node
voltage method**.

- Write the node voltage equations for this circuit. Do not solve.

- E = 200 volts, Is = 11 A, R1 = 15 Ω, R2 = 40 Ω, R3 = 10
Ω, R4 = 20 Ω, R5 = 30 Ω. Find V using the node voltage
method. The answer is an integer.

- E = 12I volts, Is = 2.5 A, R1 = 16 Ω, R2 = 6 Ω. Find I using
the node voltage method. The answer is an integer.

- E1 = 150 V, E2 = 120 V, E3 = 70 V, Is = 4I amps, R1 = 600 Ω, R2 =
17 Ω, R3 = 10 Ω, R4 = 60 Ω, R5 = 5 Ω, R6 = 100
Ω, R7 = 8 Ω. Find I using the node voltage method. The answer
is an integer.

The mesh current method simplifies and shortens the solution of problems by allowing you to write only Kirchhoff's Voltage Law. Kirchhoff's Current Law is implicit. Consider the circuit below.

Using branch currents, we can write KVL:

-V_{S} + i_{1}R_{1} + i_{3}R_{3} = 0

-i_{3}R_{3} + i_{2}R_{2} = 0

We can also write KCL at the top node, which gives:

-i_{1} + i_{2} + i_{3} = 0

Which gives:

i_{3} = i_{1} - i_{2}

Substituting into KVL gives:

-V_{S} + i_{1}R_{1} + (i_{1} -
i_{2})R_{3} = 0

(i_{2} - i_{1})R_{3} + i_{2}R_{2} =
0

With the mesh current method, we skip several steps, and just write the two
equations above. i_{1} and i_{2} are then considered to be mesh
currents rather than branch currents. Nevertheless, if an element is not shared
by two meshes, such as R_{1}, the mesh current is the same as the
branch current.

Now consider a more complicated circuit.

We want to write mesh current equations for this circuit, so we define
currents in the three meshes. The top and right meshes are labeled
I_{1} and I_{2}, respectively. The left mesh can be labeled 6
A, since we already know the current. We also define a voltage across the 4 A
source. Next, we write KVL equations for the two unknown mesh currents. The
convention for doing this is to put sources on the left, and loads on the
right.

5 - V_{I} = 2(I_{1} - 6)

V_{I} = 3(I_{2} - 6) + 7I_{2}

We need a third equation, one relating 4 A to the mesh currents. Observe
that the branch current in the 4 A source is made up of two mesh currents,
I_{2} in the direction of the 4 A current, and I_{1} opposite
to the 4 A current. Therefore:

4 = I_{2} - I_{1}

With three equations and three unknowns, we can solve for all the unknowns.

Another solution technique uses the concept of a "supermesh", and saves the writing of one equation. The circuit below is the same as before, but it is labeled differently.

The 4 A source is unlabeled (it still has a voltage, but we don't have to know what it is). The supermesh is shown by the red line. We write KVL around this supermesh as follows:

5 = 7I_{2} + 3(I_{2} - 6) + 2(I_{1} - 6) = 0

We still need the equation for the current in the 4 A source:

4 = I_{2} - I_{1}

Now we have two equations and two unknowns. Use either solution method you wish.

Before going on to the homework, you should complete Tutorial 5 on **the mesh
current method**.

- Use the mesh current method to write enough equations to solve for the
mesh currents. Do not solve.

- E1 = 195 V, E2 = 20 V, R1 = 5 Ω, R2 = 40 Ω, R3 = 20 Ω,
R4 = 50 Ω, R5 = 80 Ω. Use the mesh current method to find I.
The answer is in integer.

- Ix = 5 A, Iy = 7 A, E = 76 V, R1 = 12 Ω, R2 = 20 Ω, R3 = 8
Ω. Use the mesh current method to find V. The answer is in
integer.

- Is = 1.5I, E1 = 88 V, E2 = 52 V, R1 = 7 Ω, R2 = 3 Ω, R3 = 10
Ω, R4 = 30 Ω. Use the mesh current method to find I. The
answer is in integer.

If the circuit shown in the "black box" in Figure 1 consists only of DC voltage sources, current sources, and resistors, then the black box can be emulated by the circuit shown in Figure 2.

Figure 1. Unknown Circuit

Figure 2. Thevenin Equivalent Circuit

This means that some V_{TH} and R_{TH }can be selected that
will produce the same terminal properties as the "black box" in Figure 1.

For such a circuit, V_{TH }= V_{OC} where V_{OC} is
the open circuit voltage, the voltage at *a-b* when nothing is connected
to it. Furthermore R_{TH} = V_{OC}/I_{SC} where
I_{SC} is the short circuit current, the current when a short is
connected between *a *and *b*.

To find the Thevenin equivalent circuit of any unknown collection of sources and resistors, two measurements must be made. Assume a "black box" as in Figure 1 above. In this example we will measure the voltage across two different resistors connected to the terminals of the black box. When we connect a 100 Ω resistor to the terminals, we measure a voltage of 20 V, and when we connect a 350 Ω resistor, we get a voltage of 35 V, as shown in Figure 3 below.

Figure 3. Example Circuit for Thevenin Analysis

Kirchhoff's Voltage Law for Measurement 1 is:

-V_{TH} + V_{1} + 20 = 0

By Ohm's Law:

V_{1} = I_{1}R_{TH}

Also by Ohm's Law:

I_{1} = V/R = 20/100 = .2 A

Plugging this into KVL:

(1) -V_{TH} + .2R_{TH} + 20 = 0

Similar analysis can be done for Measurement 2:

-V_{TH} + V_{2} + 35 = 0

V_{2} = I_{2}R_{TH}

I_{2} = V/R = 35/350 = .1 A

(2) -V_{TH} + .1R_{TH} + 35 = 0

We now have two equations and two unknowns, which we can solve
simultaneously. Subtract Equation (2) from Equation (1):

.1R_{TH} - 15 = 0

(3) R_{TH} = 150 Ω

Now plug Equation (3) into Equation (1):

-V_{TH} + .2(150) + 20 = 0

V_{TH} = 50 V

The resulting Thevenin equivalent circuit is shown below in Figure 4.

Figure 4. Resulting Thevenin Equivalent Circuit

Besides being useful for characterizing unknown circuits, Thevenin's Theorem
is also good for simplifying known circuits. If we need to find the behavior
of a circuit between two points, Thevenin's Theorem can reduce the circuit to a
voltage source in series with a resistor. This is done in two steps:

(1) Find the open circuit voltage between the two points of interest. This
will be the Thevenin voltage.

(2) Short circuit all voltage sources, open circuit all current sources, and
find the resistance seen looking into the two points. This is the Thevenin
resistance.

Consider the example shown in Figure 5 below.

Figure 5. Example Circuit for Thevenin Reduction

The task is to find the Thevenin equivalent circuit between A and B. The first thing we will do is to find the open circuit voltage between A and B. Figure 6 below shows the analysis.

Figure 6. Analysis for Finding V_{OC}

Kirchhoff's Voltage Law around the dashed loop is:

-20 + V_{80} + V_{120} = 0

The Ohm's Law equations are:

V_{80} = 80I_{80}

V_{120} = 120I_{120}

These values can be plugged into the KVL equation:

-20 + 80I_{80} + 120I_{120} = 0

The Kirchhoff's Current Law equation at node C is:

-I_{80} - 1 + I_{120} = 0

Note that there is no current in the 22 Ω resistor because of the open circuit.

The KCL equation can be solved for I_{80}:

I_{80} = I_{120} - 1

Now we plug this into the KVL equation:

-20 + 80(I_{120} - 1) + 120I_{120} = 0

Then we solve for I_{120}:

-20 +80I_{120} - 80 + 120I_{120} = 0

I_{120} = (20 + 80)/(80 + 120) = 100/200 = .5 A

I_{120} can be used to solve for V_{120}:

V_{120} = 120I_{120} = 120(.5) = 60 V

Now we'll find V_{OC} using KVL:

-V_{120} + V_{22} + V_{OC} = 0

V_{OC} = V_{120} - V_{22}

Since there is no current through the 22 Ω resistor, there is no
voltage across it:

V_{OC} = V_{120} - 0 = 60 V

Now that we've found the open circuit voltage (which is the Thevenin voltage), we'll find the Thevenin resistance. Figure 7 shows the circuit with the voltage source replaced with a short circuit and the current source replaced with an open circuit.

Figure 7. Circuit for Finding R_{TH}

The 120 Ω resistor and the 80 Ω resistor are each connected
between node B and node C; therefore they are in parallel.

R_{BC} = 80(120)/(80 + 120) = 48 Ω

R_{BC} is in series with 22 Ω.

R_{TH} = R_{BC} + 22 = 48 + 22 = 70 Ω

The resulting Thevenin equivalent circuit is shown in Figure 8.

Figure 8. Result of Thevenin Circuit Reduction

Check out this Thevenin explanation and example: http://people.clarkson.edu/~svoboda/eta/cards/p12.html (Created by Dr. James A. Svoboda of Clarkson University).

Before going on, you should complete Tutorial 6 which is a
**simulated Thevenin laboaratory**.

In many applications in electrical engineering, it is desirable to draw maximum power from a circuit. It can be easily shown that this condition will exist if the load connected to the circuit is made equal to the Thevenin resistance. See Figure 9.

Figure 9. Maximum Power Transfer

P, the power in the load resistor R_{L}, will be a maximum if
R_{L }= R_{TH}.

Before going on, you should complete Tutorial 6A on **maximum
power transfer**.

Whenever an "effect" is linearly related to its "cause", then the effect due to a combination of causes is the sum of the effects due to each cause acting alone, with all other causes removed.

To remove a voltage source, replace by a short circuit.

To remove a current source, replace by an open circuit.

Example

Consider V due to the 6 V source.

By voltage divider:

V_{6} = 6×2/(2 + 12) = 12/14 = 6/7 volts

Consider V due to the current source:

By current divider:

I_{4} = 4×12/(2 + 12) = 48/14 = 24/7 amps

V_{4} = 2I_{4} = 48/7 volts

Total V: V = V_{6} - V_{4} = 6/7 - 48/7 = -42/7 = -6 V

You can't use Superposition to find power. This is because, if V = Va + Vb,
V^{2} is not Va^{2} + Vb^{2}.

Check out this superposition explanation and example: http://people.clarkson.edu/~svoboda/eta/ClickDevice/super.html (Created by Dr. James A. Svoboda of Clarkson University).

Before going on to the homework, you should complete Tutorial 6B on
**superposition**.

1. E1 = 10 V, E2 = 60 V, Is = 1 A, R1 = 120 Ω, R2 = 60 Ω, R3 = 40 Ω. Find and sketch the Thevenin equivalent circuit between a and b. The Thevenin components will be integers.

2. The unknown device consists of DC sources and resistors. When a 90
Ω resistor is connected between *c* and *d*, a voltage of
216 V results. A 15 Ω resistor connected between *c* and *d
*carries a current of 6.4 A. What current will occur if the resistor is 42
Ω? The answer is an integer.

3. E1 = 120 V, R1 = 75 Ω, R2 = 30 Ω, R3 = 70 Ω. What
value of R_{L} will cause R_{L} to consume the most power? How
much power is that? The answers are integers.

4. E1 = 660 V, E2 = 40 V, Is = 11 A, R1 = 13 Ω, R2 = 10 Ω, R3 = 20 Ω. Use the principle of superposition 3 times to find I. The answer is an integer.

5. E = 600I volts, Is = 30 mA, R1 = 1 kΩ. What value of R will cause the most power to be consumed in R? The answer is an integer.

6. BONUS (No partial credit). A cup is floating in a dishpan. A silver spoon is in the cup. The spoon is removed from the cup and dropped in the water. Does the water level in the dishpan go up, go down, or stay the same? Explain your answer.

Capacitors are usually nothing more than 2 pieces of foil (the "plates") separated from each other by an insulating material (the dielectric).

In order to save space, the foil is often wrapped up into a tube as shown. One piece of the foil is clearly on the inside while the other is on the outside. Any "noise" or interference in the environment, be it electric, magnetic, or electromagnetic, would strike the outside foil. For this reason, the outside foil is frequently grounded (or connected as electrically close to ground as possible) so that such noise will be shunted to ground and will not be present in the output of the circuit.

A rough sketch of a capacitor is shown with an indication of which lead is connected to the outside foil.

The circuit symbol for a capacitor is shown below. Note that the "passive sign convention" is used: Current is shown flowing into the positive side of the capacitor. This is the same convention as used for the resistor and inductor.

The curved side of the capacitor in the circuit symbol used to have a special meaning. It was the side to be connected to the more negative potential, as in the electrolytic capacitor described below. This special significance has been lost, and capacitors now may be shown in any orientation.

Capacitors are used in circuits to smooth voltages, reduce hum, dampen spikes, and to link circuits. For example, a capacitor might be used between two amplifier sections to block DC but allow time-varying signals to pass. Another common use of capacitors is in frequency-selection circuits (filters).

The units of capacitance are farads (F). As the farad is quite large, capacitors are frequently rated in microfarads (µF).

Capacitors are rated by capacitance and voltage rating. For example, a capacitor may be rated at 25 µF and 80 V. The voltage rating indicates the maximum voltage that the capacitor can tolerate without breaking down. There is a tradeoff between capacitance and voltage. Manufacturers can make capacitance higher by making the dielectric thinner, but that lowers the voltage rating lower.

The basic equation for the capacitor is:

i = C | dv dt |

This equation has an important implication: **You cannot change the
voltage across a capacitor instantaneously.** An abrupt change in voltage
(for example, from 3 V to 5 V) would result in an infinite value of dv/dt, and
therefore an infinite value of the current i. Since this is physically
impossible, we must conclude that the voltage on a capacitor can change only
gradually from one value to another.

Here's the equation for the capacitor in integral form:

v(t) = | 1⌠C ⌡ |
i dt + K |

K is a constant of integration. This form of the equation is usually not convenient, as it is not clear how to find the value of K. The definite integral is a more useful form:

v(t) = | t 1 ⌠C ⌡0 |
i(s) ds + V(0) |

The voltage across the capacitor at time t = 0 must be known. The integral is from s = 0 to present time t. Note that s is a dummy variable that stands in for time t in the integral; this s is a placeholder that disappears in the final result.

Another important equation for the capacitor relates voltage to charge:

Q = CV

Click this link: http://micro.magnet.fsu.edu/electromag/java/capacitor/ for a simulation on charging a capacitor.

The energy stored in a capacitor is given by the following equation:

W = CV^{2}/2

The energy is stored in the electric field that develops between the plates.

The electrolytic capacitor is a special type of capacitor. It provides more
capacitance in a given space and at a lower cost per microfarad than most
others. Unlike other types, they may only be used with substantially direct
voltage, and they must be correctly connected with regard to polarity.
Electrolytics are usually rated at rather low voltages compared to others and
have a greater leakage current than others (this means that they are less ideal
than others - resistance may become a factor in analyzing them.) Electrolytics
usually have "+++" on one end. This indicates the end that is to be connected
to the higher DC potential. **If connected backwards, it could explode**!

A new breed of capacitor is the ultra capacitor. These capacitors have extremely high capacitance. Learn more at http://www.ultracapacitors.org/index.php?option=com_content&Itemid=57&id=37&task=view.

Capacitors (even electrolytics) are very nearly ideal. This means that the capacitor equations can be trusted. A charged capacitor will leak, and the voltage across it will decay with time. Nevertheless, the leakage current is small, and the voltage will take a long time to decay.

Capacitors can be connected in series and parallel to increase voltage or
capacitance. When connected in this way, they behave *opposite* to
resistors in series parallel. For example, a 10 µF capacitor in
*parallel* with a 5 µF capacitor gives a total capacitance of 15 µF.

Inductors are basically coils of wire. The coils may be wrapped around an iron or ferrite core to better contain the magnetic flux (and increase the inductance).

The circuit symbol for an inductor is shown below. The "passive sign convention" is used: Current is shown flowing into the positive end of the inductor. This is the same convention as used for the resistor and capacitor.

Inductors are used for one or both of these properties: inductance and magnetism.

- The inductance is useful in a device called a
*choke*. It tends to prevent rapid changes in current. It is used in some power supplies. - A
*solenoid*makes use of the magnetism generated by the inductor to open or close a latch, such as a car door lock. - A
*relay*is like a solenoid, except that it uses magnetism to open or close an electrical contact. *Motors*use magnetism to generate torque for rotation.*Transformers*use inductance to couple two coils of wire so that an AC voltage can be increased or decreased.

Inductance is measured in henries (H). Many inductors have smaller inductance, measured in millihenries (mH).

The basic equation for the inductor is:

v = L | di dt |

This equation has an important implication: **You cannot change the
current through an inductor instantaneously.** An abrupt change in current
(for example, from 3 A to 0) would result in an infinite value of di/dt, and
therefore an infinite value of the voltage v. Since this is physically
impossible, we must conclude that the current through an inductor can change
only gradually from one value to another. For example, if you have current
flowing through an inductor in series with a toggle switch, and you open the
toggle switch, a high voltage will be developed across the switch contacts
causing an arc as the current rapidly decreases to zero.

Integral forms of this equation are similar to the capacitor equations:

i(t) = | 1⌠L ⌡ |
v(s) ds + K |

Again, tke K is difficult to determine, so the definite integral form is preferred:

i(t) = | t 1 ⌠L ⌡0 |
v(s) ds + I(0) |

The integral is from s = 0 to present time t. The current at time t = 0 must be known.

The energy stored in an inductor is given by the following equation:

W = I^{2}L/2

The energy is stored in the magnetic field.

Inductors are not particularly ideal. Since they are (usually) made of copper wire, they will have some resistance. For the purposes of this class, we will assume that the inductor is an ideal device, that is, that it has zero resistance.

Except for their magnetic properties, inductors are not often introduced in circuits. Capacitors are much more common.

Inductors in series and parallel are treated in exactly the same way as resistors in series and parallel.

In the circuit below, v(t) = 5 + 12sin(300t) V, L = 200 mH, and C = 25 µF. The current in the inductor is 0.15 A at t = 0. Find i as a function of time.

Note that the voltage of the source (5 + 12sin(300t) V) is also the voltage
across the inductor and the voltage across the capacitor. We can therefore
find i_{L} and i_{C} from the equations for the inductor and
capacitor. Then we will use Kirchhoff's Current Law to find i. First, let's
find the currents in the capacitor and the inductor.

i_{C} = C |
dv dt |
= 25×10^{-6} |
d dt |
[5 + 12sin(300t)] |

i_{C} = 25×10^{-6} [3600cos(300t)] = .09cos(300t) A

i_{L}(t) = |
t 1 ⌠L ⌡0 |
v(s) ds + I(0) = | t 1 ⌠.2 ⌡0 |
[5 + 12sin(300s)]ds + .15 = 5[5s - | 12 300 |
cos(300s) | t ] + .15 0 |

i_{L}(t) = 5[5t - .04cos(300t) + .04] + .15 = 25t - .2cos(300t) +
.35 A

When solving definite integrals such as the above, it's important to consider the lower limit, the evaluation of the function at s = 0.

Now we'll solve Kirchhoff's Current Law at the upper node.

-i + i_{L} + i_{C} = 0

i = 25t - .2cos(300t) + .35 + .09cos(300t) = 25t - .11cos(300t) + .35 A

Before going on to the homework, you should complete Tutorial 7 on **inductors
and capacitors**.

1. Label the diagram with currents and voltages. Use Kirchhoff's laws to
write equations to solve for the voltage across and the current through each
element. Do not solve.

2. v(t) = 5t volts, L = 50 H, C = 450 µF, R = 6 kΩ. The downward current in the inductor is 3 mA at t = 0. Find the magnitude and direction of the current flowing in all four elements as a function of time. Simplify the results.

3. i(t) = 4 + e^{-150t} amps, R = 5 Ω, L = 60 mH, C = 700 µF.
The voltage across the capacitor is 12 V at t = 0, with the left side of the
capacitor being more positive than the right. Find v as a function of time.
Simplify your answer.

4. A 500 µF capacitor has 150 V across it. If all its energy is put into kinetic energy of a .05 gram BB (a BB is a small metal pellet), how fast would the BB go?

5. The voltage on a 40 µF capacitor is found to be 12(1 - e^{-8t})
volts. Find the voltage at t = 0, and the voltage at t = 0.1 second. How much
energy is stored on the capacitor as t approaches infinity?

BONUS (No partial credit.) A candle was burning in an elevator at the top floor of a tall building. The cable snapped, and while the elevator was falling, the candle went out. What caused the candle to go out? This is not a trick question. There is a perfectly logical explanation based on physics.

Display and printer note: This lesson uses a symbol for infinity. Here it is: ∞ It should look like an 8 on its side. If it does not, consider upgrading to a more modern browser. Even if it looks correct in your browser, it may not print correctly on your printer. If it doesn't, consider upgrading your printer driver.

This lesson deals with the use of either capacitors or inductors (not both at the same time) in single-time-constant circuits. This usually means that the circuit consists of a single inductor or capacitor, several resistors, and one or more switches. Such a circuit is shown below.

Assume that the switch has been open for a long time, and that at t = 0 the switch is closed. Intuitively, you should expect that before the switch is closed, the capacitor will charge up to 50 V. This is so because, if you wait long enough, the current in the 4000 Ω resistor will decay to you, and there will be no voltage across the resistor. Then, by KVL, the capacitor and battery voltages must be equal.

After the switch is thrown, things get more complicated: The capacitor will begin to drain through the 1000 Ω resistor, but current will continue to be provided by the battery. You might guess that the capacitor voltage would decay, but wouldn't go all the way to zero, and you would be right.

Click here to see a demonstration of the RC time constant.

Below is a systematic approach to solving problems like this. Read it carefully, then we'll apply it to the above problem.

This technique works for first order systems with constant sources.

For simplicity, assume t_{i} = 0, where t_{i} is the initial
time, *i.e.* the time when the switch is closed (or opened).

1. Write down:

i_{C} = C(dv_{C}/dt)

v_{L} = L(di_{L}/dt)

i(t) = i_{f} + (i_{0} - i_{f})e^{-t/τ}

v(t) = v_{f} + (v_{0} - v_{f})e^{-t/τ}

2. For 0^{-}

- Recall that "steady state" has been reached.
- Therefore, v
_{C}= constant, i_{L}= constant - Therefore, i
_{C}= 0, v_{L}= 0 - Redraw the circuit, open circuiting capacitors (because i
_{C}= 0) and short circuiting inductors (because v_{L}= 0) - Apply KVL and KCL to find unknown voltages and currents, particularly
v
_{C}or i_{L}.

3. For 0^{+}

- Recall that you can't change v
_{C}or i_{L}instantaneously. - Therefore, v
_{C}(0^{+}) = v_{C}(0^{-}) and i_{L}(0^{+}) = i_{L}(0^{-}) - Apply KCL and KVL to find v
_{0}and i_{0}.

4. For infinity

- Repeat 2.
- Calculate v
_{f}or i_{f}.

5. Finally

- Redraw circuit, removing C or L, shorting voltage sources, and opening current sources.
- Calculate R
_{TH}"looking into" the terminals where you took out C or L. - Calculate τ = R
_{TH}C or τ = L/R_{TH}. - Plug values into i(t) or v(t) equation.
- Simplify.

1. Write down:

i_{C} = C(dv_{C}/dt)

v(t) = v_{f} + (v_{0} - v_{f})e^{-t/τ}

2. For 0^{-}

- Recall that "steady state" has been reached.
- Therefore, v
_{C}= constant - Therefore, i
_{C}= 0 - Redraw the circuit, open circuiting capacitors (because i
_{C}= 0)

- Apply KVL to find the unknown voltage.

This is easy:

-50 + v_{4000}+ v = 0

But v_{4000}= 0 because it caries no current, so

v = v(0^{-}) = 50 V

3. For 0^{+}

- Recall that you can't change v
_{C}instantaneously. - Therefore, v
_{C}(0^{+}) = v_{C}(0^{-})

v(t = 0) = v_{0}= 50 V

4. For infinity

- Repeat 2.

By the voltage divider formula,

v = 50×1000/(4000 + 1000) = 10 V

- Calculate v
_{f}.

v_{f}= v = 10 V

5.

- Redraw circuit, removing C, shorting voltage sources.

- Calculate R
_{TH}"looking into" the terminals where you took out C.

R_{TH}= 4000 × 1000/(4000 + 1000) = 800 Ω - Calculate τ = R
_{TH}C.

τ = 800 × 20×10^{-6}= 16×10^{-3 }s = 16 ms = .016 s - Plug values into v(t) equation.

v(t) = v_{f}+ (v_{0}- v_{f})e^{-t/τ}= 10 + (50 - 10)e^{-t/.016}V - Simplify.

v(t) = 10 + 40e^{-t/.016}V

Here's what a graph of the capacitor voltage looks like:

Check out this first order circuit explanation and example: http://people.clarkson.edu/~svoboda/eta/plots/FOC.html (created by James A. Svoboda of Clarkson University).

Before going on to the homework, you should complete Tutorial 8 on **first
order circuits**.

It is not necessary to use the organizational tables in any of these problems, but they may be helpful for keeping your solution neat, especially for problem 4.

1. E = 75 V, L = 40 mH, R = 22 kΩ. The switch is closed at t = 0. Find an expression for i(t) for t > 0.

Voltage or Current | Value | Reason or Calculation |

i(0^{-}) |
||

i(0^{+}) |
||

i(∞) |

2. R = 6.8 kΩ, C = 4 µF. The capacitor is initially charged to 60 V. After the switch is closed, how long will it take before the voltage has dropped to 40/e volts?

Voltage or Current | Value | Reason or Calculation |

v_{c}(0^{-}) |
||

v_{c}(0^{+}) |
||

v_{c}(∞) |

3. E = 125 V, R1 = 4.7 kΩ, R2 = 15 kΩ, R3 = 6.8 kΩ, C = 10 µF. The switch has been open for a long time. At t = 0 the switch is closed. Find v(t) for t>0.

Voltage or Current | Value | Reason or Calculation |

v(0^{-}) |
||

v(0^{+}) |
||

v(∞) |

4. E = 30 V, R1 = 33 kΩ, R2 = 22 kΩ, C = 20 µF, L = 65 mH. The
switch is closed at t = 0. Find:

i_{1}(0^{-}), i_{2}(0^{-}),
v_{1}(0^{-}), v_{2}(0^{-})

i_{1}(0^{+}), i_{2}(0^{+}),
v_{1}(0^{+}), v_{2}(0^{+})

i_{1}(∞), i_{2}(∞), v_{1}(∞),
v_{2}(∞)

Give reasons or calculations for each answer.

Voltage or Current | Value | Reason or Calculation |

i_{1}(0^{-}) |
||

v_{1}(0^{-}) |
||

i_{2}(0^{-}) |
||

v_{2}(0^{-}) |
||

i_{1}(0^{+}) |
||

v_{1}(0^{+}) |
||

i_{2}(0^{+}) |
||

v_{2}(0^{+}) |
||

i_{1}(∞) |
||

v_{1}(∞) |
||

i_{2}(∞) |
||

v_{2}(∞) |

**Bonus **(no partial credit). The 20 µF capacitor is initially
uncharged. The switch has been in position #1 for a long time. At t = 0 it is
switched to position #2.

a) What is the voltage v after a long time?

b) Find v(t) for t > 0.

Hint for a): Use conservation of charge and Kirchhoff's Voltage Law.

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In this lesson, you will study circuits that contain one capacitor, one inductor, resistors, sources, and switches. The significant difference from the lesson on first order circuits is the inclusion of both a capacitor and an inductor in the circuit. This dramatically changes the character of the resulting solutions. Although the boundary value analysis you have previously studied will still work, the solutions will typically no longer be simple exponentials. The capacitor and the inductor will exchange energy in ways that produce solutions with such components as damped sinusoids.

Consider the circuit below.

It is desired to find the voltage **v** across the capacitor
after the switch is closed. The direct approach to solving this problem would
be to write differential equations for the currents and the voltages in the
circuit, then solve them. Instead, we will solve it using a systematic approach
assuming that the solution falls within one of three general forms.

These solutions work if the circuit can be reduced to look like either of the two figures below.

Defining some variables:

α = R/2L for series circuits, or

α = 1/2RC for parallel circuits.

ω_{0}^{2} = 1/LC (resonant frequency)

x_{0} = initial value

x_{f} = final value

where x represents v (voltage) or i (current).

**Overdamped** (α > ω_{o})

x(t) = x_{f} + A_{1}e^{s1t} +
A_{2}e^{s2t}

where

s_{1} = -α + √α^{2} -
ω_{0}^{2}

s_{2} = -α - √α^{2} -
ω_{0}^{2}

**Underdamped** (α < ω_{0})

x(t) = x_{f} + B_{1}e^{-αt}cos
ω_{d}t + B_{2}e^{-αt}sin
ω_{d}t *where* ω_{d} = √ω_{0}^{2} - α^{2}

**Critically damped** (α = ω_{0})

x(t) = x_{f} + D_{1}te^{-αt} +
D_{2}e^{-αt}

Click here for a simulation of a series RLC circuit. It was created by Barry Hansen.

Use the boundary value analysis from the lesson on first order circuits to
show that:

v_{0} = 220 V

v_{f} = 0 V

Also use boundary value analysis to find v'(0) -- that's the first
derivative of v(t) evaluated at t = 0. This can be done by noting that,

i_{C} = C dv_{C}/dt

At t = 0, this becomes:

i_{C}(0) = Cv'(0)

You can use Kirchhoff's Current Law at t = 0^{+} to show that,

i_{C}(0) = 0, and therefore

v'(0) = 0

Next, we must reduce the circuit to the form of the parallel RLC circuit. This can be done with a source conversion on the voltage source and resistor.

Now we calculate two constants:

α = 1/2RC = 1/(2 × 11 × 62.5×10^{-6}) = 727 radians/second

ω_{0}^{2} = 1/LC = 1/(.025 × 62.5×10^{-6}) =
64×10^{4}

ω_{0} = 800 radians/second (Note that it's a radian frequency --
not in cycles/second.)

Since α < ω_{0}, the circuit is underdamped. We
therefore use the following equation:

x(t) = x_{f} + B_{1}e^{-αt}cos
ω_{d}t + B_{2}e^{-αt}sin
ω_{d}t *where* ω_{d} = √ω_{0}^{2} - α^{2}

Since we are looking for a voltage, this equation becomes:

v(t) = v_{f} + B_{1}e^{-αt}cos
ω_{d}t + B_{2}e^{-αt}sin
ω_{d}t *where* ω_{d} = √ω_{0}^{2} - α^{2}

Now let's evaluate some of the constants.

v_{f} = 0 V (previously calculated)

ω_{d} = √ω_{0}^{2} - α^{2} =
√800^{2} -
727^{2} = 333 rad/sec

This leaves B_{1} and B_{2} as unknowns. These must be
determined from initial conditions. At t=0,

v(0) = 220 = 0 + B_{1}e^{0}cos(0) +
B_{2}e^{0}sin(0) = B_{1}

B_{1} = 220

To evaluate B_{2}, it will now be useful to calculate v'(t), the first
derivative of v(t):

v'(t) = d/dt[v_{f} + B_{1}e^{-αt}cos
ω_{d}t + B_{2}e^{-αt}sin
ω_{d}t]

v'(t) = 0 + B_{1}e^{-αt}(-αcos ω_{d}t
- ω_{d}sin ω_{d}t) +
B_{2}e^{-αt}(-αsin ω_{d}t +
ω_{d}cos ω_{d}t)

We now evaluate the first derivative at t = 0, recalling that we previously
evaluated v'(0) to be 0:

v'(0) = 0 = B_{1}e^{0}[-αcos(0) -
ω_{d}sin(0)] + B_{2}e^{0}[-αsin(0) +
ω_{d}cos(0)]

0 = B_{1}[-α - 0] + B_{2}[0 + ω_{d}] =
220[-727] + B_{2}[333]

B_{2} = 480

The final solution is therefore:

v(t) = 220e^{-727t}cos 333t + 480e^{-727t}sin 333t V

A plot of this voltage is shown below.

Before going on to the homework, you should complete Tutorial 9 on **second
order circuits**.

1. a) Prove that this circuit is underdamped. b) An additional resistor is to be added in parallel with the 80 Ω resistor to make the circuit critically damped. What value should the resistor have?

2. E = 15 V, Is = 120 mA, R1 = 100 Ω, R2 = 330 Ω, C = 50 µF, L = 120 mH. The switch has been closed for a long time. At t = 0, the switch is opened. Find v(t) for t > 0.

3. E1 = 9 V, E2 = 45 V, R1 = 100 Ω, R2 = 150 Ω, C = 6 µF, L = 240 mH. The switch has been open for a long time. At t = 0, the switch is closed. a) Find i(t) for t > 0. b) Extra credit: Carefully plot i(t) from t = 0 to a suitable value. Use graph paper or a plotting program.

Sinusoidal currents and voltages occur frequently, and the electrical
engineer must thoroughly understand them. This requires an intimate familiarity
with the mathematical properties of sinusoids. Below is a diagram of a typical
sinusoidal voltage, given by the equation, v(t) = V_{m}cos(ωt +
φ) volts.

V_{m} is the magnitude of the voltage. ω is the angular
frequency in radians/second. φ is the phase angle in radians. Notice that
the wave will reach a peak at ωt = -φ; this point is noted in the
figure as t = -φ/ω. T is the period in seconds; in physics, this
would be known as the wave length. Period is related to frequency by the
equation,

f = 1/T.

f is the frequency in Hz or cycles/second. Naturally, the frequency f and the
angular frequency ω are related:

ω = 2πf

V_{peak-to-peak} is nothing more than what it says. Mathematically,
it's V_{max} - V_{min}, where V_{max} and
V_{min} are the maximum and minimum values of the sinusoid.

You may also have use for V_{avg}. This is the average value of the
wave. By inspection, it should be clear that the average value of the above
sinusoid is zero, because it spends as much time above zero as below zero. A DC
voltmeter connected to a sinusoidal source will read zero volts, the average
value.

Another useful parameter is the root-mean-square (rms) voltage. For all
sinusoids centered around zero, this is just the peak value divided by the
square root of 2. The rms value is also know as the *effective value*,
because a DC voltage equal to the rms value will cause the same average power
to be consumed in a resistor. An AC voltmeter connected to a sinusoidal source
will read the rms value. Because the rms value is the most useful for
sinusoidal voltages and currents, it is the default value. If a person says a
certain sinusoidal voltage is "120 volts," it is implied that this is the rms
value, not the peak value.

Preview of lessons to come: If we converted this voltage to a phasor, its
value would be V_{m}/φ. If you don't
understand what this means yet, don't panic; you soon will understand it.

Note that, in the expression cos(ωt + φ), ωt is radians/second times seconds, to give radians. Nevertheless, the angle φ is frequently given in degrees. It is therefore important to convert the units to the same type before attempting to compute the cosine. If you choose to convert φ to radians, you must also remember to configure your calculator to accept angles in radians, recalling that calculators default to assuming angles are in degrees. For this reason, most engineers convert the ωt term to degrees. This is easily done by multiplying the ωt term by 180/π, because there are π radians in 180°.

Electrical engineers prefer working with cosines, rather than sines. It's easy to convert a sine to a cosine:

sin(φ) = cos(φ - 90°)

Before going on, you should complete Tutorial 10 on
**sinusoids**.

Although the rms value of a sinusoid is the maximum value divided by the square root of 2, this is not true for other wave shapes. In general, one must calculate the root-mean-square. This is the square root of the mean of the square of the wave, as shown below.

In this equation, t_{0} is some arbitrary time, and time
t_{0} + T is the time one period later. Although the equation is
written for a voltage v, it works for current i also. The wave must be
periodic (repetitive), or it is not possible to find an rms value.

Before going on, you should complete Tutorial 10A on **root
mean square**.

Electrical engineers deal with many sinusoidal problems by converting sinusoidal currents and voltages to phasors. Phasors allow the engineer to replace difficult-to-solve time-domain differential equations with frequency-domain phasor equations, which are easier to solve. Phasors make use of complex numbers. Before studying phasors, we must therefore first understand how to manipulate complex numbers. You probably already learned how to do this in math courses, so the material below is just a review, with special emphasis on the way electrical engineers use complex numbers.

Although most mathematicians use *i* to represent the square root of
minus one, electrical engineers use *j*, to avoid confusion with their
usual variable name for current.

j = √-1

j × j = -1

1/j = -j

**M** = general form of a complex number, written in bold face,
or with an overline: M

Me^{jθ} = exponential form, where M is the magnitude and θ
is the phase angle, usually in radians

M/θ = polar form, where M is the magnitude and
θ is the phase angle, usually in degrees

a + jb = rectangular form, where a = Mcosθ, and b = Msinθ

(a , b) = rectangular form, as shown on many calculators

(a / b) = polar form, as shown on many
calculators

You must become adept at converting from one form to another. Most
scientific calculators have provisions for converting from rectangular to polar
coordinates, and *vice versa*. If your calculator cannot do that, then
you must use the sine, cosine, and tangent functions to do conversions. Take
care when using the tangent and arctangent functions. The complex number -5 +
j5 converts to 7.07/135°, not 7.07/-45°. The arctangent cannot distinguish between (-5)/5
and 5/(-5).

You must also be accomplished at doing math with complex numbers. Again, advanced scientific calculators can do math using complex numbers. If your calculator isn't that sophisticated, you'll have to make frequent use of these three important mathematical relationships:

(a + jb) + (c + jd) = a + c + j(b + d)

(a/λ)(b/φ) =
ab/(λ + φ)

(a/λ)/(b/φ) =
(a/b)/(λ - φ)

These relationships show that addition and subtraction are easy in rectangular coordinates, but multiplication and division are easy in polar coordinates. You should therefore convert complex numbers to rectangular form if you intend to add or subtract, and to polar form if you intend to multiply or divide.

Many calculators, even so-called "scientific" calculators, do not have adequate complex number features for electrical engineering use. If a calculator has "rectangular to polar conversion," such as the TI-82, this is helpful, but is far from ideal. When purchasing a calculator, it is wise to get one that has "complex number" capability, that is, a calculator able to directly handle mathematical operations such as (5 + j6)x(7/18°). The TI-83, -85, and -86 have this capability, as does the HPGX.

Below are some links that will help you learn how to use your calculator to do complex math:

Soon (in the next lesson), you will apply what you have learned about sinusoids and complex numbers to the solution of circuits in "the phasor domain." To do this, you must know how to convert currents and voltages in the time domain to their equivalents in the phasor domain. Although you will learn more about this in the next lesson, the basics are given below.

To convert a sinusoidal time-domain voltage or current to a phasor, drop the cosine and the ωt, and use only the magnitude and the phase angle. Example:

250 cos(65t + 73°) volts transforms to 250/73° volts.

To convert a phasor to a sinusoid, just reverse this process. Note that you must know the angular frequency to do the inverse transform. In the example below, assume the frequency is 50 Hz.

345/-81° amps transforms back to 345 cos(2π50 - 81°) amps = 345 cos(314t - 81°) amps.

Before going on to the homework, you should complete Tutorial 10B on
**complex numbers**.

1. A certain voltage is found to be 50sin(65t + 23°) volts. Find angular frequency in radians per second, frequency in Hz, period, maximum voltage, minimum voltage, peak-to-peak voltage, rms voltage, average voltage, voltage expressed as a phasor, the average power consumed by a 22 Ω resistor having this voltage, and the voltage at t = .02 s.

2. Convert the following complex numbers to polar form.

10 + j20, 6 + j89, 12 - j18, (-44) + j16.

3. Convert the following complex numbers to rectangular form.

5/11°, 110/-58°, 20/140°, 120/-175°

4. Evaluate these expressions. Give answers in both rectangular and polar form.

(510/64°)/(45 + j19), 18 + j22 - 19/23°

5. If the frequency is 60 Hz, what is the time function corresponding to a phasor of 90 + j70 volts?

6. A sinusoidal voltage has a maximum amplitude of 12 volts and a period of 17 ms. At t = 0, its value is 8.4 volts and is rising. Write an expression for this voltage as a function of time. Note: Do not ignore "and is rising."

7. A 440 V, 60 Hz sinusoidal source is connected to a 100 Ω load. What is the rms current, peak current, and average power consumed by the load?

8. Find the average and rms voltages of this periodic wave.

9. Find expressions for M and θ in symbolic form.

M/θ = (A/λ)(B/φ)/(C/α)

10. Find X and Y. Use your calculator (not Maple, Mathematica, or another math program). Note that X and Y are complex numbers; you must therefore solve a system of complex equations. Give your answer in rectangular form. X and Y must satisfy both equations. Also note that since X and Y are complex numbers, each has a real and imaginary part. Hint: In the solution for these equations, both the real and imaginary parts of X and Y are non-zero.

25X + j10Y = 5/26°

(13 - j9)X + 6Y
= 3 + j4

BONUS (No partial credit

**Find the Flaw in the Proof**

Given: t > 3

3t > 9

3t - t^{2} > 9 - t^{2}

t(3 - t) > (3 - t)(3 + t)

t > 3 + t

0 > 3

When you finish this lesson, you should be able to:

- Use Kirchhoff's Laws for phasors.
- Write phasor mesh current and node voltage equations and solve them.
- Combine complex impedances in series and parallel.
- Find complete solutions of problems in the time domain.

The problems in this unit are rather special. They all have sinusoidal
forcing functions. They all involve only the steady state solution. This may
seem rather restrictive, but a vast number of problems in electricity fall into
this category. Almost all problems in AC circuits do. The power of the phasor
method lies in the fact that when a sinusoidal forcing function is used in a
linear system, the results (voltage and current) are *always*
sinusoidal, and of the same frequency as the forcing function. Thus, if a 120
V, 60 Hz voltage is used to drive a system of resistors, inductors and
capacitors, the steady-state (what happens after waiting a long time) current
in the system will be a 60 Hz sinusoid. Its amplitude and phase angle are the
only unknowns that must be calculated.

Click here here to see how an AC generator works.

Using the phasor method to get a complete solution to a problem requires 3 steps:

- Transformation of the system from time functions into phasors and complex
numbers. Notice that
*the resultant system is independent of time*. - Solution of the problem using complex numbers. Kirchhoff's Laws, node voltage method, mesh current method, and combining of impedances in series and parallel may all be useful in doing this.
- Transformation of the system back into the time domain.

This method may at first seem long and tedious, but it is far easier than solving the original time domain problem using differential equations.

The mesh current and node voltage methods can be used for AC circuits by using the same methods as with DC circuits, with the addition of impedance concepts and phasor voltages and currents.

Try this site for a good explanation of phasors: http://people.clarkson.edu/~svoboda/eta/phasors/Phasor10.html (created by Dr. James A. Svoboda of Clarkson University).

- Transform sources into phasors using only magnitude and phase angle;
*e.g.*100cos(200t + 3°) V transforms to 100/3° V. - Transform inductors by converting L to jωL. The units of jωL are ohms.
- Transform capacitors by converting C to 1/jωC. The units of 1/jωC are ohms.
- Resistors do not require transformation.
- Transform unknown functions of time (variables) by writing them as
general phasors (using boldface or overline);
*e.g.*v_{2}(t) transforms to V_{2}.

Below is an example of a network and its transform.

Continuing with the same example, I can be
calculated as,

I = 85/27°/(-j88.9 + 110
+ j135) = 85/27°/(110 + j46.1) = 85/27°/119.3/22.7° = 0.712/4.3° A

An inverse transform can then be done on I to
get:

i(t) = 0.712cos(450t + 4.3°) A

V_{I} - V =
R_{1}(**I** - I_{3}) +
(1/jωC)(I - I_{2})

V = (1/jωC)(I_{2} - I) +
jωL(I_{2} - I_{3}) + I_{2}R_{2}

0 = I_{3}R_{3} + jωL(I_{3} - I_{2}) + R_{1}(I_{3} - I)

This is 3 equations and 3 unknowns. The first equation is not required unless it is necessary to find the voltage across the current source. If only the mesh currents are needed, equations 2 and 3 are sufficient.

Node A: -I + (V_{A} - V_{C})/R_{3} + (V_{A} - V_{B})/R_{1} = 0

Node B: (V_{B} - V_{A})/R_{1} + (V_{B} - V)/(1/jωC) + (V_{B} - V_{C})/(jωL) = 0

Node C: (V_{C} - V_{A})/R_{3} + (V_{C} - V_{B})/(jωL) + V_{C}/R_{2} = 0

The complex version of resistance is called *impedance*. It is a
combination of *resistance* and *reactance*. Resistance is the real
part of the impedance, and reactance is the magnitude of the imaginary part.
It's expressed this way:

Z = R + jX

R = resistance, the real part of impedance

X = reactance, the imaginary part, without the j.

The reactance for an inductor:

X_{L} = ωL

The units of X_{L} are ohms: radians/second × henries is ohms.

The reactance for a capacitor:

X_{C} = -1/ωC

Note the minus sign. X is real, not imaginary. You have to add the j to
make it imaginary. The units of X_{C} are ohms: radians/second ×
farads is inverse ohms. Capacitors always have negative reactance and negative
impedance.

To convert reactance into equivalent impedance, just add the j:

Z_{L} = jX_{L} = jωL

Z_{C} = jX_{C} = -j/ωC

Ohm's Law has a complex version:

V = I Z

It's the same a Ohm's Law for resistors, except that all the numbers are complex. The analysis for series/parallel combinations of impedances is the same as it is for resistances.

Before going on to the homework, you should complete Tutorial 11 on
**phasors**.

1. v(t) = 240sin(400t - 52°) volts, i(t) = 8cos(400t + 110°) amps, R = 100 Ω, L = 220 mH, C = 12 µF. Transform this network into the frequency domain.

2. V**s** = 72/55°
V, I**s** = 5/-29° A,
R = 17 Ω, Z** _{L}** = j11 Ω,
Z

3. V**s** = 7/33°
V, I**s** = 4/-179°
A, R1 = 175 Ω, R2 = 80 Ω, R3 = 217 Ω, Z** _{L}** = j101 Ω, Z

4. R = 15 Ω, L = 24 mH, C = 9 µF. Find the impedance between **a**
and **b** at a frequency of 750 Hz. Give your answer in polar
coordinates.

5. v_{s}(t) = 100cos(2200t - 33°) volts, R = 22 Ω, L = 24 mH,
C = 150 µF. Find v(t) (steady state).

**Bonus **(no partial credit). This problem will test your
ability to apply a concept we learned earlier (superposition) to the solution
of a phasor problem with two sinusoidal sources. It is crucial to note that
the two sources have different frequencies. Find v(t) (steady state).

You have learned the following equation for DC circuits:

P = VI

A more general equation for all circuits is this one:

p(t) = v(t) × i(t)

At any instant in time, this equation will be true. Nevertheless, engineers are usually more interested in average power than in instantaneous power.

P = I^{2}R and P = V^{2}/R are well known equations for
power consumption in DC circuits. Similar equations apply for AC circuits:

P = I_{rms}^{2}R

P = V_{rms}^{2}/R

P in the above two equations refers to average power. The rms subscript is
usually left off so that the equations P = I^{2}R and P =
V^{2}/R are understood to use rms values. This will be a general rule:
When I and V are used in an AC circuit, they are meant to be rms values.

Similarly, an AC circuit can consume "reactive power," given by these
equations:

Q = I^{2}X

Q = V^{2}/X

X is the reactance, as in Z = R + jX. Unlike R,
X can take on negative values. So, although P (real power) is always positive,
Q (reactive power) may be positive or negative. The reactive power is not in
the units of watts, because it is not real power. The units are *VARs*,
which stands for *volt-amps reactive*. Real power represents the power
available to do work. Reactive power can never do any work -- it's just a
measure of energy storage in the circuit.

In some books, you will find Q replaced by P_{x} for reactive
power.

In a general AC circuit, the angles of the phasor voltage and current will
be different; *e.g.*

V = 120/32° V

I = 5/95° A

It will be useful to find the difference between these two angles:

θ = θ_{v} - θ_{i}

θ is called the *power factor angle*.

In this specific case,

θ = 32° - 95° = -63°

Since the voltage and current angles are different, it should not be
surprising that the power P is not just the product of V (the rms voltage) and
I (the rms current). In fact, this is the relationship:

P = EIcosθ watts

In this equation, E is the same as V, the rms voltage. The cosθ part
of the equation is referred to as the *power factor*. We usually work
with the power factor rather than the angles themselves.

For the example above,

P = EIcosθ = (120)(5)cos(-63°) = 272 W

Power factor = .45

In manufacturers' literature, power factor is frequently expressed as a percentage. In the above case, the power factor would be given as 45%.

The power factor may be *leading* or *lagging*. Recall that
cosθ = cos(-θ). How do we differentiate between the two? When the
current lags the voltage, as it does in an inductor, we get a positive value
for θ, and we say that the power factor is lagging. When the current
leads the voltage, as it does in a capacitor, we get a negative value for
θ, and we say that the power factor is leading. Sometimes power
engineers speak of inductors absorbing magnetizing VARs and capacitors
providing magnetizing VARs.

Here's a mnemonic for remembering the leading/lagging relationships for
inductors and capacitors:

ELI the ICE man.

The ELI part implies voltage leads current in an inductor. The ICE part implies
current leads voltage in a capacitor.

If the load is primarily inductive, Q will be positive. Most motors have lagging power factors because they are made of wire and iron, just like inductors.

L and C modify instantaneous power but do not use any average power. Their presence can, nevertheless, change the amount of average (real) power consumed in a circuit.

1 Horsepower = 745.7 watts.

Motors and generators are not 100% efficient. In a motor, for example, the input electrical power is always reduced by windage, friction, and ohmic loss before it can appear as output mechanical power. If a motor has an efficiency of 89%, for example, and puts out 5 HP (mechanical), the input power (electrical) is 5 × 745.7/.89 = 4189 W.

Apparent power is defined as:

S = EI

E and I are rms values of voltage and current. S is called the apparent power because it is the power that would result if the power factor angle were zero. The units are volt-amps (not watts). We now have three sets of units: watts for real power, VARs for reactive power, and volt-amps for a combination of the two.

The relationship among P, Q, and S is best understood with the
* power triangle*, shown below.

The relationships are just what you'd expect from a triangle:

S^{2} = P^{2} + Q^{2}

P = Scosθ

Q = Ssinθ

This triangle clearly shows why apparent power (S) is not the same as real power. You will use this triangle repeatedly to solve power problems.

Real power in AC circuits is given by these equations:

P = EIcosθ watts

=I^{2}R

=V_{R}^{2}/R

where cosθ is the power factor (pf).

Reactive power in AC circuits is given by these equations:

Q = EIsinθ VARs

= I^{2}X

= V_{x}^{2}/X

= Ptanθ

X is positive for inductors and negative for capacitors.

kVA = kilo-volt-amps

By conservation of energy, total power consumed in a circuit is equal to the sum of the power consumed by each element. Also, power generated equals power consumed. Likewise, reactive power generated equals reactive power consumed.

Advice: ALWAYS DRAW A WELL-LABELED DIAGRAM SHOWING THE POWER TRIANGLES(S).

In the example below, we consider the case of an AC generator feeding a motor through a transmission line.

One of the first things to notice is that the motor is not specified in terms of its impedance, but rather in the way it uses power. This is a 20 kW motor. This means that it uses 20 kW of real power. It has a power factor of 0.6. We could use this to calculate θ, but we will see that finding the angle is not necessary.

The task is to find the motor current, then find the voltage, power, and power factor of the generator.

We begin with a power triangle:

Here's how we calculate the triangle values:

S = P/pf = 20×10^{3}/.6 = 33.3 kVA

Q = √S^{2} -
P^{2}

Recalling that S = EI, we can find I, the motor current:

I = S/E = 33.3 × 10^{3}/120 = 278 A

The motor current is the same as the transmission line current, so we can
make the following calculations for losses in the transmission line:

P = I^{2}R = 3.86 kW

Q = I^{2}X = 23.19 kVAR

Now we will use conservation of energy to determine the net power loss in
the motor/transmission line combination:

P_{T }= P_{motor} + P_{line} = 20 kW + 3.86 kW = 23.86
kW

Q_{T} = Q_{motor} + Q_{line} = 26.7 kVAR + 23.19 kVAR =
48.89 kVAR

With these two values, we can now draw another power triangle for the whole circuit.

S is found from S^{2} = P^{2} + Q^{2}. The power
factor is just P/S:

pf = P/S = 23.86/55.3 = .431 lagging

We know the power factor is lagging, because the load is inductive.

Finally, we'll find the generator voltage:

E_{g} = S/I = 55.3×10^{3}/278 = 199 volts

In the above example, a 22 kVAR capacitor is placed across the motor to
partially correct the pf. Find the new values of E_{g}, pf, and P
necessary for the generator.

Before beginning this solution, notice that the capacitor was specified in
kVAR, not in microfarads. This is common practice in industrial applications.
Catalogs of industrial capacitors will specify the kVAR and voltage rating of
the capacitor. Recall that Q is positive for inductors and negative for
capacitors. Nevertheless, the capacitor is specified as 22 kVAR and not -22
kVAR. This is because the engineer using industrial catalogs is expected to
*know* that Q is negative for capacitors, and therefore using the minus
sign is superfluous. You, the student, will also be expected to know this. In
the diagram below, the actual (negative) value is used. This may not always be
the case with other diagrams.

From Example 1, we know that for the motor,

Q = +26.7 kVAR

Therefore the total Q for the motor and capacitor together is,

Q = 26.7 - 22 = 4.7 kVAR

This results in the following power triangle:

Notice that the power of the motor did not change as a result of adding the
capacitor. Now, we'll calculate the current in the line:

I = S/E = 20.54×10^{3}/120 = 171 amps

Before going on, let's note that the line current (171 A) is less than the motor current (278 A). This occurs because the capacitor has corrected the power factor to be more nearly unity.

For the transmission line:

P = I^{2}R = 1.47 kW

Q = I^{2}X = 8.79 kVAR

Now we apply conservation of energy to get the totals:

P_{T} = 20 kW + 1.47 kW = 21.5 kW

Q_{T} = 4.7 + 8.79 = 13.5 kVAR

This allows us to draw the composite power triangle.

pf = P/S = 21.5/25.4 = 0.85 lagging

E_{g} = S/I = 25.4×10^{3}/171 = 149 volts

Notice that addition of the capacitor has allowed us to reduce the generator voltage. This is a direct result of improving the power factor.

Before going on, you should complete Tutorial 12 on **AC
power**.

It's simple for residential users of electric power. They pay about 10 cents per kW-h. Note that this is an energy charge, not a power charge. In a few markets, residential users may be charged at a different rate dependent on the time of day; electricity may be cheaper at night.

Commercial users, such as banks and shops, may pay a lower energy charge (perhaps 4 cents/kW-h) than residential users, but they may also pay a base service charge plus a maximum load charge. They may also be required to keep the power factor above some specified minimum.

Industrial customers pay the lowest energy charge of all, perhaps 3 cents/kW-h or less. Instead they may pay in many other ways:

**Base Service Charge:** This is a basic monthly fee of perhaps $300.

**Demand Charge:** This is a fee based on the maximum amount of power (not
energy) that is used during the month. The power a customer uses varies from
time to time during the day and from day to day. The power company monitors
this power use and notes the peak power used during the billing month. For
example, the industrial customer may have had a maximum power usage of 300 kW;
at the rate of (say) $8/kW, the customer would pay $300 × 8 = $2400 as a
demand charge. Usually, the demand charge varies throughout the year, with the
highest demand charge occurring during the summer months.

**Power Factor:** The power company rewards industrial customers who have
good power factors (PF near 1), and punishes those who have poor power factors
(PF << 1). This carrot/stick approach is usually applied as an
adjustment to the demand charge. For example, a customer might have a lower
demand charge rate if his power factor is above 80% (PF = .80), and a higher
demand charge if it's below 80%.

We learned earlier that we could maximize the power consumed in a resistive
load if we set the load resistance equal to the Thevenin resistance of the
circuit the load was connected to, (*i.e.* R_{L} =
R_{TH}). For complex circuits, the result is similar:

Z** _{L}** = Z

The asterisk (*) indicates *complex conjugate*. To find the complex
conjugate of a complex number, change the sign of the imaginary part (if in
rectangular form) or change the sign of the angle (if in polar form). Below
are two examples.

(5 + j6)* = (5 - j6)

20/-35° * = 20/35°

Consider the circuit below.

For Z** _{L}** to consume the most
power, it must have an impedance of (8 - j7)* = 8 + j7 Ω.

Now let's calculate the power consumed in Z** _{L}**. Note that the total impedance seen
by the voltage source is (8 - j7) + (8 + j7) = 16 Ω . The impedance seen
by the voltage source is totally real; the imaginary parts cancel out. We can
now calculate the current:

I = 100/25° /16 = 6.25/25° A

To calculate the power consumed in the load, we just want the magnitude of
I (*i.e.* 6.25 A). Recall that phasor
currents are based on the *maximum* value of the current, not the
*rms* value. To calculate power, we need the rms value:

I_{rms} = I_{m}/√2 = 6.25/√2 = 4.42 A

We then calculate the power in the load as follows:

P = I_{rms}^{2}R_{L} = (4.42)^{2} × 8 =
156.25 W

Note that we use only the real part of the load impedance in this
calculation. We do not use the imaginary part of **Z _{L}** as this
would give us reactive power.

Before going on to the homework, you should complete Tutorial 12A on
**maximum power transfer in complex circuits**.

1. In Example 1, what kVAR value of capacitance would be needed to exactly correct the power factor of the load (the motor and capacitor together) to unity? What must the voltage of the generator be so that the motor voltage (with the capacitor in place) is 120 V?

2. Find the real and reactive generator power. The motor puts out 3 HP and has an efficiency of 89.5%.

3. Two machines and a lighting load are connected in parallel to a 220 V, 60 Hz line. The first machine uses 40 kW of power at 0.81 power factor lagging. The second machine uses 27 kVA at 0.93 power factor leading. The lighting load consumes 13 kW. What is the current in the line feeding the three loads? What kVAR value of capacitance connected across the line is necessary to obtain a power factor of 0.98 lagging? With this capacitor installed, what will be the current flowing in the line feeding the four loads?

4. **E** = 100/15°, R = 70 Ω,
**Z _{1}** = j330 Ω,

BONUS (No Partial credit) *It is said that a friend once gave Norbert
Wiener this problem. Dr. Wiener is the famous mathematical physicist who
invented cybernetics, and his friend wanted to see if Dr. Wiener thought like a
physicist or like a mathematician. * A garbage truck starts out 5 miles from
the city landfill. The truck travels at 10 mph. A fly on the garbage truck
takes off and heads for the dump at 20 mph. When the fly gets to the dump, it
turns around and heads back to the truck. Arriving at the truck, the fly again
turns around and heads for the dump. The fly repeats this procedure over and
over again until the truck arrives at the dump. How far does the fly go (the
total distance, so the answer is not "5 miles")?

Operational amplifiers are used frequently in many analog circuits, such as signal amplifiers, active filters, signal conditioners, and impedance matchers. The circuit symbol is shown in Figure 1.

Figure 1. Operational amplifier circuit symbol

Although the inputs may be shown in either orientation, it is usually more convenient to show the inverting input on the top.

The operational amplifier (op amp) is usually supplied as a small integrated circuit with one or two op amps on the chip. In addition to the signals shown in Figure 1, a practical op amp usually also has "offset null" connections for fine tuning.

The V^{+} and V^{- }supply connections are required. They
provide power for the op amp and determine the maximum and minimum excursions
of the output voltage. As long as these excursions are not exceeded,
V^{+} and V^{-} need not be considered when making op amp
calculations. Based on this concept, a simplified op amp circuit can be used,
as shown in Figure 2.

Figure 2. Simplified operational amplifier circuit symbol

Figure 3 shows a circuit model of the op amp. v_{n} is the voltage
between the inverting input and ground. v_{p} is the voltage between
the noninverting input and ground.

Figure 3. Operational amplifier circuit model

R_{i} is the input resistance (more generally, the input
*impedance *would be used) of the op amp. Typically, this resistance is
very large - usually in the millions of ohms. Therefore, the current flowing
through it will be very small - usually a few microamps or even nanoamps.

The voltage-controlled voltage source in the circuit amplifies the
difference voltage v_{d}, which is v_{p} - v_{n}, by
a factor A. This A is called the gain of the op amp, and it is usually very
large; a typical value for A is one million. Later on, we will discuss the
gain of the circuit. Do not confuse the *gain of the op amp* with the
*gain of the circuit*.

R_{o} is the output resistance (more generally, the output
*impedance *would be used) of the op amp. This will be relatively small,
perhaps 100 Ω. The output voltage v_{o} is taken with respect to
ground.

Let us use this model of the op amp to calculate the output voltage for the circuit in Figure 4.

Figure 4. Sample op amp circuit.

Let us suppose that this is a very poor op amp. Its input impedance is rather low at 18 kΩ, its output impedance is rather high at 1 kΩ, and its gain is only 2000. The result, using our circuit model, is shown in Figure 5.

Figure 5. Sample op amp circuit using op amp model.

Let's solve for v_{n} using the node voltage method:

(v_{n} - 1)/900 + (v_{n} + 2000v_{n})/(9000 + 1000)
+ v_{n}/18000 = 0

The first term represents the current through the 900 Ω resistor. The
second term represents the current from v_{n} to the dependent source
by way of the 9 kΩ resistor and the 1 kΩ resistor; note that the 9
kΩ resistor and the 1 kΩ resistor can be considered in series if no
current is drawn from v_{o}; also note that v_{d} =
v_{p} - v_{d} = -v_{n}. The third term represents the
current through the 18 kΩ resistor.

To solve this equation, first multiply by 18000:

20v_{n} - 20 + 1.8v_{n} + 3600v_{n} + v_{n}
= 0

v_{n} = 0.00552 V

This is approximately 6 mV, which is very nearly zero. It is negligibly small.

This should lead to a very small current in R_{i}. It does:

I_{Ri} = .00552/18000 = 0.307 μA

This current is certainly negligibly small.

Now let's calculate the current in the 900 Ω resistor:

I_{900} = (1 - .00552)/900 = 1.1050 mA

The current in the 9 kΩ resistor is

I_{9k} = (.00552 + 2000 × .00552)/(9000 + 1000) = 1.1046 mA

Notice that these two currents are very nearly the same. For practical purposes, they can be assumed to be equal.

Now let's calculate v_{o}:

v_{o} = v_{n} - v_{9k} = .00552 - 9000I_{9k}
= .00552 - 9000 × .0011046 = -9.94 V

This voltage is very nearly -V_{s}R_{f}/R_{s}, which
calculates to be -10 V.

Let us summarize: Even for this very poor op amp,

v_{n} is approximately 0.

I_{Ri} is approximately 0.

I_{Rs} is approximately I_{Rf}.

v_{o} is approximately -V_{s}R_{f}/R_{s}.

Since these relationships are nearly true, even for this very poor op amp, we can guess what the relationships will be for an ideal op amp circuit:

v_{n} = 0.

I_{Ri} = 0.

I_{Rs} = I_{Rf}.

v_{o} = -V_{s}R_{f}/R_{s}.

R_{i} = infinity. Therefore I_{i} = 0 at either the n or p
input.

R_{o} = 0.

A = infinity. Therefore, v_{d} = 0, (v_{p} - v_{n}) =
0, and v_{p} = v_{n}.

Using this ideal model for the op amp will greatly simplify our calculations, and will result in only very small errors in resulting voltages and currents.

The op amp is very rarely used "open loop" to amplify a tiny signal into a
big one. Instead, feedback is used. The effect is to make the circuit
insensitive to variations in A, R_{i}, and R_{o}, even rather
large variations. This makes it possible for the engineer to design op amp
circuits (almost) without regard to which brand of op amp is selected.

The following analysis will work for almost all op-amp circuits. Even if the op-amp cannot be considered ideal, this works as a good approximation.

**1) Calculate v _{p}**. This is the voltage at the positive
input. To make this calculation, assume that no current flows into the
positive input.

We will now solve an op-amp circuit using the ideal op-amp analysis method. Consider Figure 6 below.

Figure 6. Example circuit using ideal op-amp model.

1) Since no current flows into the positive input, no current flows in the
8.9 kΩ resistor. Therefore, the voltage on both sides of the 8.9
kΩ resistor is the same (1 V). So v_{p} = 1 V.

2) v_{n} = v_{p} = 1 V.

3) KCL at v_{n} is:

-I_{S} - I_{F} = 0

To find the I values, use this form of Ohm's Law:

**I _{XY} = (V_{X} - V_{Y})/R**

This important equation says that the current between points X and Y is the voltage on the X side minus the voltage on the Y side divided by the resistance. Applying this to the two currents in the circuit, we get:

I

I

Plugging these into KCL gives:

-2/10 k - (v

Multiplying by 80 k:

-16 - v

v

You can use this method to analyze almost any op-amp circuit.

There is a reason for the presence of the 8.9 kΩ resistor. It is the parallel combination of the resistors connected to the negative input. Real op-amp circuits have tiny "bias currents" that flow into the positive and negative inputs. By balancing the resistances seen by these two terminals, the effects of the bias currents tend to cancel out.

Before going on, you should complete Tutorial 13A on
**operational amplifiers**.

Most electric power is transmitted over high voltage three-phase lines. This method of power transmission is more efficient than single phase power, as you might use in your home. As you might expect, three-phase power transmission uses three wires. The voltage between any wire and ground has the same magnitude as the voltage between any other wire and ground, but the phases are different. Each voltage will be 120° out of phase with the other voltages, as shown in the figure below.

Each color represents a different phase. At any point in time, the sum of the three voltages is exactly zero.

Three-phase power is produced by specially designed three-phase electric generators. When a wire moves through a magnetic field, a voltage is generated in the wire (Faraday's Law). For a three-phase generator, three coils of wire are placed at angles of 120° to each other, and are rotated in a magnetic field 60 times per second. This produces three voltasges which are 60 Hz sinusoids, each out of phase with the others by 120°.

There are two ways these three coils of wire can be connected together. The two connections are called delta and Y. For a delta-connected generator the coils are connected together as shown in the figure below.

The coils of wire shown do not represent inductors; rather they represent
windings of the generator. Each such winding generates what is called a
*phase voltage*, meaning the voltage of one phase of the source, shown
in the figure as V_{φ}. Each winding also has a *phase
current*, I_{φ}. The phase voltage and phase current will
produce a *line voltage*, V_{L}, and a *line current*,
I_{L}.

To back up a bit, this discussion covers only *balanced
three-phase*. This means that all loads for all phases are exactly the
same. In practice, this is (usually) very nearly true. The analysis of
unbalanced three-phase is much more difficult and is beyond the scope of this
course. Given that this is balanced three-phase, each V_{φ} is
exactly the same as every other, keeping in mind that these variables represent
rms magnitudes. The same is true for I_{φ}, V_{L}, and
I_{L}.

To return to the delta-connected generator, you can easily see that any pair
of lines is directly connected to one phase of the generator. It should
therefore be obvious that:

(1) V_{L} = V_{φ}

The relationship between I_{L} and I_{φ} is not nearly
so obvious. One must sum the phasor currents in two phases of the source to
find the line current. The result is shown below.

(2) I_{L} = √3I_{φ}

This peculiar result occurs because the currents are 120° out of phase with each other.

A Y connected generator has coils connected as shown in the figure below.

It should be clear from this figure that the current in any phase must be
the same as the current in its connected line.

I_{L} = I_{φ}

The line voltage can be found by taking the phasor sum of the voltage in two
phases of the source. The result is shown below.

(4) V_{L} = √3V_{φ}

The analysis of three-phase circuits reduces to little more than deciding where to put the square root of three.

The loads for three-phase power can also be either delta- or Y-connected, as shown in the figure below.

The voltage across each **Z** is a phase voltage,
V_{φ}, for the load. We can reuse Equations (1) and (2) above for
the delta-connected load. We can reuse Equations (3) and (4) for the
Y-connected load. The same equations apply for sources and loads. For
example, the current in a Y-connected load is the same as the line current.

The analysis of a three-phase circuit is virtually the same as the analysis for a single phase AC circuit. We can still use phasor analysis, power triangles, and all the rest. There are a few specialized equations for total and phase power that may be helpful. These equations work for both delta and Y loads.

The power consumed in one phase of the load can be found if the current
resistance of the phase are known:

(5) P_{φ} = I_{φ}^{2}R_{φ}

By conservation of energy, the total power consumed by the load must be 3
times the power consumed in any phase:

(6) P_{T} = 3P_{φ}

To get the total power consumed in the load, the following equation is
useful:

(7) P_{T} = √3V_{L}I_{L}cosθ where cosθ is
the power factor.

The voltage and currents in the above equations are rms values, not peak values.

In the circuit below, a three-phase Y-connected generator is connected through power lines to a delta-connected load. Each phase of the generator produces 277 V. The load consumes 180 kW of power with a power factor of 66% lagging. The task is to find the following: (a) the line voltage, (b) the phase voltage of the load, (c) the line current, (d) the phase current of the generator, (e) the phase current of the load, and (f) the power consumed by one phase of the load.

(a) From Equation (4) for Y-connected sources or loads:

V_{L} = √3V_{φ} = 480 V

(b) From Equation (1) for delta-connected sources or loads:

V_{φLoad} = V_{L} = 480 V

(c) From Equation (7):

P_{T} = √3V_{L}I_{L}cosθ

I_{L} = P_{T}/(√3V_{L}cosθ)

I_{L} = 180,000/(√3 × 480 × 0.66) = 328
A

(d) From Equation (3) for Y-connected sources or loads:

I_{φGen} = I_{L} = 328 A

(e) From Equation (2) for delta-connected sources or loads:

I_{φLoad} = I_{L}/√3 =
328/√3 = 189 A

(f) From Equation (6):

P_{φ} = P_{T}/3 = 180 kW/3 = 60 kW

Most commercial power plants use Y-connected generators with the center tap grounded. If the phases are balanced (all phases have exactly the same current), no current will flow through the grounded center tap. When single-phase power is needed from a three-phase system, it can be extracted from any one of the three phases.

Most large motors are three-phase delta-connected motors. Three-phase motors have a distinct advantage over single-phase motors: the power is constant. In a single-phase motor, p(t) is a sinusoid, but in a three-phase motor, p(t) is a constant.

The line voltage and line current, V_{L} and I_{L}, are more
useful than the phase voltage and phase current, V_{φ} and
I_{φ}. This is because V_{L} and I_{L} are much
easier to measure. V_{φ} and I_{φ} can only be
measured by opening up the machine.

Most utilities will provide a customer with only one type of electrical service, either single-phase or three-phase. (See http://www.federalpacific.com/university/transbasics/chapter6.html and http://en.wikipedia.org/wiki/Three-phase_electric_power for further explanation). If the system is single-phase, it will probably be three-wire 120 V and 240 V. For three-phase systems, more options may be available: 240 V three-wire, 480 V three-wire, 600 V three-wire, 208Y/120 V four-wire, 480Y/277 V four-wire.

Power is measured in three-phase systems using the "two-wattmeter method," as shown below.

The two wattmeters are connected to any two phases of the input, and then to
the load as shown. Each wattmeter measures the average power, the average of
v(t) × i(t), not V_{rms} × I_{rms}. Wattmeter 1 senses the
current in the top wire and the voltage between the top wire and the center
wire; it calculates the average of the product of the voltage and the current.
If Wattmeter 1 measures P_{1} and Wattmeter 2 measures P_{2},
the total power and total reactive power can be calculated:

P_{T} = P_{1} + P_{2}

Q_{T} = √3(P_{1} -
P_{2})

Before going on to the homework, you should complete Tutorial 13 on **three
phase**.

1. Use a circuit similar to Figure 4 above to design an op amp circuit that will have an output of -20 V. Available devices: an ideal op amp, a 1 V battery, a 2 V battery, and the following resistors: 1 kΩ, 2 kΩ, 4 kΩ, 10 kΩ, 20 kΩ, 40 kΩ. Show the calculations for your design and sketch your design.

2.

Carefully show your analysis of the circuit, using the assumptions for an
ideal op amp. Find v_{o}.

3.

Carefully show your analysis of the circuit, using the assumptions for an
ideal op amp. Find v_{o}.

4.

Each V_{s} in the above figure has an rms value 220 V. Each one is
120° out of phase with the other two. R = 90 Ω, **Z _{L}** =
j110 Ω. Find the magnitude (rms value) of the current in any line and
the total power consumed by the load.

5. A Y-connected three-phase load has a phase voltage of 440 V and a phase current of 15 A. Each phase consumes 5.2 kW of power. What is the line voltage, the line current, and the total power consumed by the load? What is the power factor of the load?

6. A delta-connected three-phase source has a phase voltage of 220 V. It feeds a delta-connected load consisting of three identical impedances of 10/42° Ω each. Find the magnitude of the phase current in the load, the line current, and the total power consumed by the load.

7. A balanced Y-connected three-phase motor uses 20 kW of power at 0.81 pf lagging when connected to 220 V lines. What is the line current? What is the current in one phase of a delta-connected generator?

**Bonus** (no partial credit). In Problem 6, solve if each wire in the
transmission line has a resistance of 1 Ω. (Hint: Use delta-Y
transformation.)

**Major Topics: **electrical safety, effects of electricity on
the body, fibrillation, electrical resistance of the body, residential
wiring.

**Minor Topics:** skin puncture, body resistivity, lightning,
GFCIs, isolation transformer, defibrillator.

**Objectives:** After studying this lesson you should be able
to:

1. Determine the probable effect of an electric current between two points on the body.

2. Determine current level in the body.

3. Wire up a wall socket.

4. Explain how a GFCI works.

5. Answer questions on electrical safety.

Muscles and nerves of the body conduct information by electrical and electrochemical means. For example, electrical impulses generated in the brain travel down the spinal cord, are transmitted to peripheral nerves, and then activate muscle movement. Likewise, external stimuli, for example touch or heat, generate electrical signals that travel through nerves to the brain. It is reasonable to expect, therefore, that electricity applied externally to the body will affect nerves and muscles.

Table 1 presents data on the effects of 60 Hz current on the body. This frequency is used as an example, because 60 Hz current is the most common source of current and the most common cause of accidental electrocution. It should be understood that the values given here are approximate, and that authorities and experts in the field are in disagreement over some of the values.

Current |
Effect |

20 A | Permanent brain damage |

5 A | Respiratory arrest |

2 A | Central nervous system damage |

1 A | Burns |

80 ma | Ventricular fibrillation |

50 ma | Asphyxia |

9 ma | Muscles frozen |

1 ma | Pain |

0.2 ma | Threshold of perception |

0 | No effect |

**Table 1.** Effects of continuous 60 Hz current

between the two arms. Currents given are approximate.

Starting at the bottom of the table, the threshold of perception is 0.2 ma, which means that currents below this level cannot be detected. Currents near this level are not dangerous in themselves, but they may be dangerous in a contributory sense. For example, if a man standing on a tall ladder received a low-level shock, he might be startled into falling off.

Currents above 1 ma are often painful, but really dangerous currents are those above 9 ma, which can cause the muscles of the body to be frozen. This occurs because the externally applied electricity overrides the internally generated signals to the muscles, making them impossible to move. More precisely, this "let-go-current" threshold is defined as 9 ma for men (½ of 1% of men cannot let go of a 9 ma current) and 6 ma for women. This situation is obviously extremely dangerous, and it is made worse by the fact that skin resistance decreases the longer the current is on. Therefore, if the voltage is constant, the current will increase the longer the person is connected to it.

Asphyxia occurs when the chest muscles are frozen, making it impossible for
the person to breathe. The value of 50 ma given for this is somewhat fuzzy
since values in the literature range from 18 ma to 60 ma. Recall that the
current was defined as flowing between the two arms, so if the entrance and
exit points for current are different from this, one can expect a different
value for the asphyxia threshold. The chest has a larger cross-sectional area
than the arm, so there will be a higher current density
(amps/meter^{2}) in the arm than in the chest. This can be seen in
Figure 1, and it is the reason that 9 ma of current may cause the arm muscles
to freeze but not the chest muscles. If a person is attached to a current that
causes asphyxia, he will die if he is not removed from the current within about
5 minutes.

Figure 1. Current flow in the body.

The principal cause of death by accidental electrocution is from ventricular fibrillation. The heart, like any other muscle of the body, is affected by electrical currents, but the heart has some special electrical properties: it is synchronous and self-stimulating. This means that the heart will continue to beat even when disconnected from the nervous system, and that it will beat in a rhythm which efficiently pumps blood. Ventricular fibrillation is asynchronous contraction of the heart muscle. When this occurs blood flow essentially stops. This can occur from a number of causes, most of which involve "heart failure" of various kinds, but we are concerned here with fibrillation caused by external electrical stimulation. If an electrical current passes through the heart it my "depolarize" some areas of the heart while leaving others unaffected. This can interrupt the synchrony of the heart and cause it to go into fibrillation. Fibrillation remains even after the electrical stimulation has ceased. External cardiac massage may be of some use until (and if) the heart naturally regains its synchronous rhythm, but a more effective method is to employ a defibrillator.

Most defibrillators found in hospitals consist of two large electrodes and a capacitive discharge circuit. The two electrodes are placed on the chest, and a large capacitor charged with a high DC voltage is discharged through the chest. The high current causes the heart to completely depolarize. When the current is removed, the heart should start to beat properly. To reemphasize an earlier point, it is partial depolarization (electrical excitation) that causes fibrillation; complete depolarization allows the heart to regain its synchrony.

If fibrillation is not corrected within about 5 minutes, the victim will die. Permanent brain damage due to oxygen starvation may occur within 3 minutes. This is the reason why heart teams in hospitals are required to react with speed and efficiency to a cardiac arrest. With external heart massage, medication, and resuscitation, the times mentioned above can be extended to varying degrees.

A 60 Hz current between the two arms of 80 ma would cause ½ of 1% of the
population to go into ventricular fibrillation. More precisely,

I_{fib} = K/√T

where K = 100 for children and 185 for men, and T is time in seconds of
exposure to the shock. This equation is fairly accurate for 8.3ms < T < 5
sec. If a shock is directly applied to the heart, such as may accidentally
happen with a defective cardiac catheter, as little as 80µa can cause
fibrillation.

Above currents of 1 A, skin and subcutaneous burns occur. At a current of 2 A central nervous system (CNS) damage may occur. This current level may damage the spinal cord and cause temporary or permanent loss of feeling in the area through which the current flowed.

Currents above 5 A between the arms may cause respiratory arrest. Naturally, currents flowing in the head, rather than between the arms can cause respiratory arrest at much lower currents than 5 A (approximately 100 ma). Unlike asphyxia, which is caused by the freezing of the chest muscles, respiratory arrest continues after the current is removed. The victim stops breathing because the "breathing center" in the medulla has been disrupted. Mouth-to-mouth resuscitation should be administered to a victim who is suffering from respiratory arrest until (and if) the CNS recovers from the shock. Recovery may not occur for up to ½ hour.

Currents greater than 20 A result in permanent brain and nerve damage by
causing excessive heat generation (due to I^{2}R). The spinal cord and
lower brain would be damaged the worst, possibly resulting in paralysis from
the neck down.

Although current is the direct hazard in accidental electrocution, it is voltage that causes current to flow. This is simply a statement of Ohm's Law. If we know the resistance of the body and the applied voltage, we can calculate the current that will flow.

Body resistance, excluding the skin, is approximately 500 Ω between
any two points. More exact values can be calculated knowing that the
resistivity, ρ, of the body is about 1 Ω-m and using the formula,

where A is cross-sectional area of the tissue, and S is the distance the
current travels through the tissue. This equation is seldom necessary, since
skin resistance is usually much more than internal body resistance.

In general, the total resistance of the skin at the point of entrance and
exit of the current is between 1 kΩ and 100 kΩ. The former applies
if the areas are wet and large, the latter if the skin areas are dry and small.
More precisely, the resistance of a l cm^{2} area of skin is
approximately 200 kΩ ± 100 kΩ when dry, and 1% of that when wet. As
mentioned previously, skin resistance drops when current flows through it for a
long time.

Some calculations can be made using the above information. If a person were accidentally connected to at 120 V 60 Hz source (common residential power), typical current flow for dry skin would be,

I = 120 V/(100 kΩ + 500 Ω) = 1.2 ma

Checking this against Table 1, we see that the current is painful, but not particularly dangerous.

If the skin is wet:

I = 120 V/(1 kΩ + 500 Ω) = 80 ma

This is quite possibly a lethal current.

Based on considerations such as this, and actual case histories, it can be shown that if the voltage is less than about 40 volts, one is relatively safe from electrocution. This is not to say that one could not imagine a situation where 40 V could be lethal; but there are no recorded electrocutions at voltages this small.

On the other hand, if the voltage is extremely high, there may be somewhat less risk of death than for a moderately high voltage. This surprising assertion is because at very high voltages the heart may completely depolarize and therefore not fibrillate.

If the voltage or frequency is high, skin resistance becomes negligible. For V > 240 volts, "skin puncture" may occur -- a hole is burned in the skin. For f > 1000 Hz, the skin behaves like a capacitor rather than a resistor, and the reactance becomes small even though the resistance may be high. Reactance is the imaginary part of impedance and is similar to resistance in many ways. The reactance of a capacitor is inversely proportional to frequency. Therefore, as the frequency increases, the current through a capacitor will increase.

DC currents are less dangerous than 60 Hz currents of equivalent rms value because the DC current does not change with time. The 60 Hz current does, and therefore presents greater danger of fibrillation by presenting various currents to the heart. On the other hand, very high frequencies (above about 10 kHz) are not particularly dangerous since the current changes directions so rapidly that muscle and nerve cells do not have time to depolarize. Indeed, high frequency (500 kHz) and high voltage (800 V) power is used in "electrosurgery" -- the use of electricity to cut tissue and coagulate bleeding tissue. It turns out that frequencies between 60 Hz and 400 Hz (common commercial power frequencies) are the most dangerous of all.

Electrical accidents account for about 1000 deaths per year in the United States. Lightning accounts for about 100 more deaths and about 300 injuries. Almost all of the lightning deaths occur between the months of May and September. This is probably because most of the thunderstorms occur in this period and also because people are more likely to be outdoors.

Click here to see a lightning simulation.

About 10% of the lightning deaths are tree related. Lightning will tend to strike the tallest object in the vicinity. Thus it would be safe to take shelter under a tree in a forest, but hazardous to do the same thing on a golf course.

About 2% of the lightning deaths are phone related. Lightning can travel down phone lines and enter residences. Lightning arresters (see Figure 2) are provided to prevent this.

Figure 2. Lightning arresters.

Lightning arresters are designed to short high voltage to ground, but occasionally they fail.

Figure 3 was prepared by Dr. Donald J. Scheer, and is used with permission. It shows the way three-phase power is converted to voltages suitable for use in the home. Three-phase power arrives as three wires strung at the top of the utility pole. The voltage between any pair of wires is 12,400 V, and the voltage from any wire to ground is 7,200 V. One of the overhead three-phase wires is tied through the primary coil of the distribution transformer to ground. The secondary of the transformer is center-tapped to provide two 120 V single-phase supplies. The two ends of the secondary of the transformer provide 240 V single-phase. This "service drop" may serve as many as 15 residential customers.

Figure 3. Conversion of three-phase power to single-phase power.

Figure 4 shows the wiring diagram for electricity entering a residence. The watt-hour meter on the outside of the house measures the amount of electrical energy that is used. From the service entrance box, the breaker panel or fuse box distributes 120 V and 240 V power for use in the residence. In older homes, fuses may be used instead of circuit breakers.

Figure 4. House wiring block diagram.

To carry power throughout a building, conduit is probably the best installation to have. Nevertheless, few homes are equipped with conduit. It is found much more often in industrial buildings. Conduit is thin-walled metal pipe that can be bent with a special tool. It carries the black wire (120 V, "hot") and the white wire (120 V return, at ground potential). It also acts as a ground for all "boxes" and other electrical connections in the house. This ground ordinarily carries no current; that is the job of the white wire. Conduit provides good mechanical protection and fire protection for the wiring.

Next best, depending on the installation, is armored cable, commonly called "BX." This is a flexible spiral-wound metal cable that (usually) carries the black wire, the white wire and a bare ground wire. It provides fairly good protection for the wiring. It is not waterproof.

The most common (and cheapest) installation in residential use is plastic
coated cable, officially NM cable, but usually called "romex," although
Romex^{®} is a registered trademark of General Cable Industries. Romex
comes in 2-wire and 3-wire varieties. The 2-wire variety has no ground wire,
and is not approved for new installations. NM cable provides minimal mechanical
and fire protection, but it is waterproof (in the NMC version), so it is
suitable for underground use. Three-wire "romex" is acceptable for residential
wiring under Kentucky and Louisville wiring codes. Romex is the most common
cable found in residential wiring.

Figure 5 shows a detail view of an outlet box. The box itself may be metal or plastic. Metal boxes are themselves grounded. Note how the hot (black), return (white), and ground (green or bare) wires are connected.

Figure 5. Typical outlet box connection details.

Figure 6 shows various kinds of wall sockets. There are many other kinds of sockets, but these four are the most common in residential wiring. When correctly wired, the 120 V black wire should be on the right and the white should be on the left. Occasionally 3 prong safety sockets are installed with the ground on top, thus making the return on the right. Most older installations have standard or polarized sockets. New installations in Kentucky must have 3 prong safety sockets.

Figure 6. Various wall sockets.

The polarized socket is intended to allow certain types of plugs (with one prong larger than the other) to be inserted in only one way, so that the correct side of the equipment will be grounded. Some television sets have such plugs. Three prong sockets will accept obsolete, polarized or 3 prong plugs, and these are considered the safest and best sockets.

**Procedure for replacing a receptacle.**

- Turn off power to the socket either at the breaker box or at the wall (if the socket is controlled by a wall switch).
- Use a multimeter to test to be sure the power is off.
- Remove the face plate.
- Remove the screws connecting the receptacle to the box.
- Remove the old receptacle.
- Connect the
**black**wire to the**brass**screw of the new receptacle. - Connect the
**white**wire to the**chrome**screw. - Connect the
**green**or**bare**wire to the**green**screw. - Screw the new receptacle into the box.
- Replace the face plate.
- Use the multimeter to do a continuity check to assure that there is zero resistance between the return line and the ground line.
- Turn the power on.
- Use the multimeter to check the voltage between the hot side and the return side of the receptacle.

3 prong plugs are the safest because the third wire is provided to ground the chassis of the equipment. Figure 7 shows a properly grounded appliance. If an internal fault should occur that accidentally connects the 120 V "hot" side to the chassis, the fuse will blow or the circuit breaker will open. The user will not be hurt. Figure 8 shows an ungrounded appliance connected to the power source with a 2 prong plug. An internal fault may cause 120 V to be connected directly to the chassis. This is very dangerous and potentially lethal.

Figure 7. Properly grounded appliance.

Figure 8. Ungrounded Appliance.

When attempting to connect 3 prong plugs to 2 prong sockets, a dangerous device known as an adapter or "cheater" is often used. This is shown in Figure 9. If this device is used at all, it should be used properly, that is by connecting the green wire under the screw on the outlet.

Figure 9. 3 prong to 2 prong adapter.

Another method of making safe electrical equipment is by "double insulation." In such equipment, the case (chassis) is usually nonconducting and the internal parts are insulated. Double insulation has a good safety record, but it is still dangerous when wet.

Most electrical accidents occur by interposing a person as a path to ground (not to return) for 120 V. A device has been invented that can sense this occurrence and correct it. It is the ground fault circuit interrupter (GFCI) shown in Figure 10. This is sometimes just called a "ground fault interrupter" (GFI). The current flowing in the black and white wires is the same under ordinary circumstances, thus the flux in the toroid is zero. If a ground fault occurs, the current in the black wire will exceed the current in the white wire, flux will be developed in the toroid and a voltage in the coil. This will activate the control circuit which will open the relay shutting off the power and saving the victim from electrocution. Ground fault trip current is approximately 5 ma. GFCIs are now required on all outdoor electrical installations and in all wet locations, such as kitchens and bathrooms. When installed in indoor locations, GFCIs are sometimes daisy-chained with ordinary receptacles. This is done to save money. If properly wired, all sockets on the daisy chain have ground fault protection.

If provided for all electrical installations, GFCIs would eliminate 81% of electrocutions and many electrically caused fires. Double insulation, if universally used, would eliminate 57% of electrocutions.

Figure 10. Ground fault circuit interrupter.

Another method of protection from shock is the isolation transformer shown in Figure 11. If either side of the output is touched (but not both simultaneously) there is no danger because there is no complete circuit to ground.

Figure 11. Isolation transformer.

- When changing fuses (or working on any circuit you suspect may be "hot") keep one hand in your pocket.
- When working on circuits, remove jewelry, including rings, necklaces, bracelets, and watches with metal watch bands.
- Use the "buddy system." Have somebody stand by the breaker box to make sure nobody turns it back on while you are working.
- Never use a drill (or other power tool) with the third prong removed.
- Especially when working outdoors, make sure your tools are properly grounded.
- Do not overload circuits.
- Beware of appliances with metal cabinets if you have wet hands or feet.
- Don't put extension cords under rugs.

You are standing barefoot on a wet concrete floor. Which points are safe to touch (i.e. non-lethal) and why?

Work these out for yourself, then click to see the correct answers.